Shu wrote: Find all polynomials $P(x)$ with real coefficients such that \[xP\bigg(\frac{y}{x}\bigg)+yP\bigg(\frac{x}{y}\bigg)=x+y\] for all nonzero real numbers $x$ and $y$. $p(x)=ax+b$ with $a+b=1$.

With $x,y:=1$, we see that $P(1)=1$. Now, let $Q(t):=P(t)-(1-c)t-c$, where $c:=P(0)$. We see that $Q(0)=0$, $Q(1)=0$, and $xQ\left(\frac{y}{x}\right)+yQ\left(\frac{x}{y}\right)=0$ for all $x,y\neq 0$. Replacing $y$ by $1$, we obtain $Q(x)=-x\cdot Q\left(\frac{1}{x}\right)$, showing that $\deg(Q) \leq 1$. As $Q(0)=Q(1)=0$, we conclude that $Q \equiv 0$, so $P(t)=(1-c)t+c$.

Put $y=1$ to get $xP\left({1\over x}\right)=x+1-P(x)$ As the RHS contains no negative powers of $x$, all of those contained in $P\left({1\over x}\right)$ must be canceled by the factor $x$. Thus $P$ is linear. Put $P(x)=ax+b$ into the initial equation and simplify to get $b=1-a\implies P(x)=ax-a+1,a\in\mathbb{R}$.

let $y=kx,$then$p(k)+kp(\frac{1}{k})=k+1$ let $q(k)=p(k)-1$,then$q(k)=-kq(\frac{1}{k}),q(1)=0$ let$q(k)=(k-1)t(k)$ then$t(k)=t(\frac{1}{k})$if $deg(t(k))>0$, when $k\rightarrow$+∞,$LHS\rightarrow$∞,$RHS\rightarrow t(0)$,contradiction. hence $t(k)=c,p(k)=c(k-1)+1$. QED