This is essentially an application of the division algorithm, but the details require significant care.

View $A(x,y),B(x,y)$ as polynomials in $x$. Then, use division algorithm to get \[A(x,y)=Q(x,y)B(x,y)+R(x,y)\]where $\deg_x R<\deg_x B$. Here $R,Q$ are polynomials in $x$ with coefficients being rational functions in $y$. Thus, $B(x,y)\mid R(x,y)$ for infinitely many values of $y$. But this means that $\deg B(x,y)<\deg_x B(x,y)$ for infinitely many values of $y$, so the leading coefficient of $B(x,y)$, which is a polynomial in $y$ is $0$ for infinitely many values of $x$. This is a contradiction, so $R(x,y)=0$. Therefore, $A(x,y)/B(x,y)=Q(x,y)$ where $Q(x,y)$ is a polynomial in $x$ with coefficients being rational polynomials in $y$. So we have \[Q(x,y)=x^n\frac{Q_n(y)}{P_n(y)}+\cdots+\frac{Q_0(y)}{P_0(y)}.\]By doing polynomial division on each coefficient, we can write $Q(x,y)=\bar{Q}(x,y)+C(x,y)$ for some polynomial $C$, and \[\bar{Q}(x,y)=x^n\frac{R_n(y)}{P_n(y)}+\cdots+\frac{R_0(y)}{P_0(y)}\]where $\deg R_i<\deg P_i$ for all $i$. Letting $P=\mathrm{lcm}(P_n,\ldots,P_0)$, we get that \[\bar{Q}(x,y)=\frac{x^nR_n(y)\frac{P(y)}{P_n(y)}+\cdots+R_0(y)\frac{P(y)}{P_0(y)}}{P(y)}.\]We have that the degree of the top is now strictly less than that of the bottom, so this cannot be a polynomial in $y$ for infinitely many $x$, unless the top is $0$. So $\bar{Q}=0$, so $A=B\cdot C$, as desired.

Claim: We can write $A(x,y)/B(x,y)=Q(x,y)$ where $Q(x,y)$ can be written as a polynomial in $x$ with coefficients rational functions in $y$. Proof: For some infinite set of reals $S$, we have for any $y_0\in S$ that $B(x,y_0)\mid A(x,y_0)$ as polynomials in $x$. Treat $A$ and $B$ as polynomials in $x$ with coefficients polynomials in $y$; call $A$ and $B$ as $A(x)$ and $B(x)$ in this framework. By the Division Algorithm we can write \[ A(x) = B(x)Q(x) + R(x), \qquad (\spadesuit)\]for some polynomials in $x$ (technically $x$ and $y$) $Q(x)$ and $R(x)$ which have coefficients rational functions in $y$. By putting the sum $R(x)$ over a common denominator, we see that we can write $R(x)$ as \[ R(x) = \frac{C(x,y)}{D(y)} \]for some $C(x,y) \in \mathbb{R}[x,y]$ and $D(y) \in \mathbb{R}[y]$. For an arbitrary $y_0\in S$, plug in $y=y_0$ into $(\spadesuit)$. Since $B(x,y_0)\mid A(x,y_0)$ as polynomials in $x$, we have $R(x,y_0)=0$. So for infinitely many $y_0$, we have $R(x,y_0)=0$. Note that $D(y)$ can only have finitely many roots, so when we write $C(x,y)$ as a polynomial in $y$ with coefficients polynomials in $x$, there exists an infinite set of values for which plugging $y$ into $C(x,y)$ outputs 0. This implies $C(x,y)=0$ identically, so $R(x)=0$. Hence $A(x)=B(x)Q(x)$, proving the claim. $\blacksquare$ Now by the claim, \[ \frac{A(x,y)}{B(x,y)} = \frac{F_n(y)}{G_n(y)}x^n + \frac{F_{n-1}(y)}{G_{n-1}(y)}x^{n-1} + \cdots + \frac{F_0(y)}{G_0(y)}x^0\]for some $F_i,G_i \in \mathbb{R}[y]$. Use polynomial division to write $F_i(y) = (--)G_i(y)+G_i'(y)$ for some polynomials $(--) \in \mathbb{R}[y]$ and $G_i(y) \in \mathbb{R}[y]$ with $\deg G_i'(y) < \deg G_i(y)$ (this is not the derivative). Moving all the $(--)x^i$ terms to the left side, we have \[ \frac{A(x,y)}{B(x,y)} - C(x,y) = \frac{G_n'(y)}{G_n(y)}x^n +\cdots + \frac{G_0'(y)}{G_0(y)}x^0 \]for some $C(x,y) \in \mathbb{R}[x,y]$. Hence the RHS must be a polynomial in $\mathbb{R}[x,y]$, but the degree of $y$ when we put the sum over a common denominator is less on the numerator than the denominator, so it cannot be a nonzero polynomial in $\mathbb{R}[x,y]$. Hence it is 0, so \[ \frac{A(x,y)}{B(x,y)} - C(x,y) = 0,\]finishing.

The idea is that we abuse division algorithm. First of all, we can think of $A$ and $B$ as polynomials in $x$, where each coefficient is a polynomial in $y$. As a result, we can do \begin{align*} A(x, y) = B(x, y) Q(x, y) + R(x, y), \end{align*}where the $x$-degree of $R$ is less than the $x$-degree of $B$, but the coefficients of $R$ are all rational functions in $y$, and similarly $Q$'s coefficients are rational functions in $y$. Dividing, we get \begin{align*} \frac{A(x, y)}{B(x, y)} = Q(x, y) + \frac{R(x, y)}{B(x, y)}, \end{align*}and by picking arbitrarily large $y$ we make it so that the highest $x$ degree terms of $R(x, y)$ and $B(x, y)$ are both nonzero. However, since $\deg_x R < \deg_x B$, this is a contradiction and thus $R$ is the zero polynomial for all arbitrarily large $y$. This implies that the rational function coefficients of $R$ are identically $0$. Let the degree of $Q$ be $q$, and define $a, b$ analogously. Then we can write \begin{align*} Q(x, y) &= x^{q} \cdot \frac{S_q(y)}{T_q(y)} + \cdots + x \cdot \frac{S_1(y)}{T_1(y)} + \frac{S_0(y)}{T_0(y)} \\ &= y^p \cdot \frac{M_p(x)}{N_p(x)} + \cdots + y \cdot \frac{M_1(x)}{N_1(x)} + \frac{M_0(x)}{N_0(x)}, \end{align*}where the second equality is from doing the exact same thing but thinking in terms of polynomial in $y$. Denote $N(x) = \text{lcm}(N_0(x), \cdots, N_p(x))$, and define $T(y)$ similarly. Now we find that \begin{align*} N(x) \cdot (x^qS_q(y) \cdots \text{some polynomial}) = T(y) \cdot (y^p M_p(x) \cdots \text{some other poly}). \end{align*}However, since $\gcd(N(x), T(y)) = 1$, we see that $N(x)$ divides the $y^pM_p(x) \cdots$ polynomial, but this is bad since the two are relatively prime by assumption. Thus we are done.

By division algorithm, we may write \[A(x,y)=Q(x,y)\cdot B(x,y)+R(x,y),\]where \(Q\) and \(R\) are polynomials in \(x\) (and rational functions in \(y\)) and such that \(\deg_x R<\deg_x B\) (meaning \(R\) has smaller degree than \(B\) when considered as polynomials in \(x\)). We are given that for infinitely many \(y_0\), we have \(B(x,y_0)\) divides \(R(x,y_0)\) (as polynomials in \(x\)), implying \(R(x,y_0)=0\). This implies that \(R(x,y)\equiv0\), so \[Q(x,y)=\frac{A(x,y)}{B(x,y)}\]is a polynomial in \(x\). Symmetrically, \(Q(x,y)\) is a polynomial in \(y\). This means we can find polynomials \(F\), \(U\), \(G\), \(V\) such that \[Q(x,y)=\frac{F(x,y)}{U(y)}=\frac{G(x,y)}{V(x)}.\]Assume since \(\mathbb R[x,y]\) is a UFD that \(\deg\gcd(F,U)=\deg\gcd(G,V)=0\). Evidently \(\deg\gcd(U,V)=0\) as well, so \(U\) is coprime to both \(F\) and \(V\). However \(G(x,y)\cdot U(y)=F(x,y)\cdot V(x)\), so \(U\) must divide \(F\cdot V\). This implies \(\deg U=\deg V=0\), implying the conclusion.

From the problem statement, we know that there exists two infinite real sequences $\{x_i\}_{i=1}^{\infty}$, $\{y_j\}_{j=1}^{\infty}$, such that $\frac{A(x_i,y)}{B(x_i,y)}$ is a polynomial in $y$, and $\frac{A(x,y_j)}{B_(x,y_j)}$ is a polynomial in $x$. Suppose that $$\frac{A(x_i,y)}{B(x_i,y)}=R_d^{(i)}y^d+R_{d-1}^{(i)}y^{d-1}+\dots+R_0^{(i)}$$for $i=1,2,\dots,N$, where $N$ is an integer larger than $\deg_x A$. By Lagrange Interpolation Theorem, there exist polynomials $F_k\in \mathbb{R}[x]$ with degree lower than $N$, passing through points $$(x_1,R_k^{(1)}),(x_2,R_k^{(2)}),\dots,(x_N,R_k^{(N)})$$for each $0\leq k\leq d$. Consider the polynomial $$S(x,y)=A(x,y)-B(x,y)\sum\limits_{k=0}^dF_k(x)y^k$$Note that $\frac{S(x,y_j)}{B(x,y_j)}=\frac{A(x,y_j)}{B(x,y_j)}-\sum\limits_{k=0}^dF_k(x)y_j^k$ is a polynomial in $x$ with degree less than $N$ for every $j\geq 1$, and it has $x_1,x_2,\dots,x_N$ as its zeros. Thus it must be a zero polynomial, meaning that $\frac{S(x,y_j)}{B(x,y_j)}=0$ for any $x\in \mathbb{R}$. We conclude that $S(x,y)\in \mathbb{R}[X,Y]$ is a zero polynomial. Hence take $C(x,y)=\sum\limits_{k=0}^dF_k(x)y^k$, we have $A=B\cdot C$.

View $A(x,y)$ and $B(x,y)$ as polynomials in $x$. By the division algorithm, we can write $$\frac{A(x,y)}{B(x,y)}=Q(x,y)+\frac{R(x,y)}{B(x,y)}$$where $Q$ and $R$ are polynomials in $x$ with coefficients as rational polynomials in $y$, and $\deg_x R<\deg_x B$. Since the RHS should be a polynomial in $x$ for infinitely many values of $y$, we may pick some value of $y$ which makes all the rational polynomial coefficients (other than those that are identically zero) nonzero and well-defined. Then it is clear that $R(x,y)$ should be identically zero in $x$, but this implies that $R(x,y) \equiv 0$ in both $x$ and $y$, i.e. $\frac{A(x,y)}{B(x,y)}$ is a polynomial in $x$ with coefficients as rational polynomials in $y$. By a symmetrical argument, it is also a polynomial in $y$ with coefficients as rational polynomials in $x$. Therefore we can write $$\frac{A(x,y)}{B(x,y)}=\frac{F(x,y)}{G(y)}$$where $F$ and $G$ are polynomials. By polynomial division we can again write the RHS as $S(x,y)+\frac{T(x,y)}{G(y)}$ where $\deg_y T<\deg_y G$, where $S$ is a polynomial in both $x$ and $y$, so $\frac{T(x,y)}{G(y)}$ should also be a polynomial in $y$ (with coefficients as rational polynomials in $x$), but this is impossible because $\deg_y T<\deg_y G$, unless $T \equiv 0$, which would imply that $C \equiv S$ works. $\blacksquare$

Solved with GrantStar. Bad problem Also might be a fakesolve We solve the problem over $\mathbb C [x,y]$. It's left as an excercise to verify that it's impossible for $C$ to have complex coefficients while $A$ and $B$ have real coefficients. Claim: The quotient $A(x,y)/B(x,y)$ is a polynomial in $x$ for all $y$. Proof: Treating $y$ as a constant, compute the single-variable polynomial division $A(x,y)/B(x,y)$ with respect to $x$. The quotient and remainder will both be polynomials in $x$ with coefficients that are rational functions in $y$. We're given that the remainder $R(x)$ is the zero polynomial in $x$ for infinitely many $y$. So, each coefficient of $R(x)$ (a polynomial in $y$) is equal to $0$ for infinitely many $y$, meaning every coefficient of $R(x)$ is identically zero. In particular, there are finitely many values of $y$ which result in division by zero errors so they are negligible. So, for cofinite $y$, $A(x,y)/B(x,y)$ is a polynomial in $x$. This means that for cofinite $y$, the quotient $A(x,y)/B(x,y)$ doesn't equal infinity for any $x$ – similarly, for cofinite $x$, the quotient doesn't equal infinity for any $y$. So, there are a finite number of points at which the quotient could equal infinity. If there was a factor of $B(x,y)$ that didn't divide $A(x,y)$, this factor would equal $0$ for infinitely many pairs $(x,y) \in \mathbb C^2$, contradiction. So, $B(x,y)$ divides $A(x,y)$.