The answer is $f \equiv cx$ for all $x \in \mathbb{R^+}$ and some constant $c$. Verifying is easy. Let $P(x, y)$ be the assertion above. $\textbf{Claim: }$ $f$ is strictly increasing, hence injective. $\textit{Proof}$ Trivial as $(x, x+y^2f(x^2))$ can attain any two pairs of values in $\mathbb{R^+}$. $\square$ Let $f(1) = c$. $P(1, 1) \implies f(c+1) = c(c+1)$. $P(1, c+1) \implies f(c^3+2c^2+c+1) = c^4+2c^3+c^2+c$ Notice this gives us that there exists $a > b$ where $\frac{a}{f(a)} = \frac{b}{f(b)}$. Now then $P(\sqrt{a}, \sqrt{\frac{y}{f(a)}}) \implies f(y+\sqrt{a}) = f(\sqrt{a}) + f(\sqrt{y}\sqrt{\frac{a}{f(a)}})^2$, $P(\sqrt{b}, \sqrt{\frac{y}{f(b)}}) \implies f(y+\sqrt{b}) = f(\sqrt{b}) + f(\sqrt{y}\sqrt{\frac{b}{f(b)}})^2$ Taking the difference means $f(y+\sqrt{a}) - f(y+\sqrt{b}) = f(\sqrt{a}) - f(\sqrt{b}) \implies f(y+p) = f(y) + C$ for all $y \geq N$. $P(x, \sqrt{\frac{p}{f(x^2)}}) \implies C = f(\sqrt{p}\sqrt{\frac{x^2}{f(x^2)}})^2$. Since $f$ is injective, we have that $\frac{x^2}{f(x^2)}$ needs to be constant, which means $f \equiv cx$ for all $x \geq N$. Now taking $y$ small and $x$ arbitrarily large easily gives $f \equiv cx, \forall x \in \mathbb{R^+}$.

The answer is $f\equiv Vx$ forall $x\in\mathbb{R^+}$ for some constant positive $V$. Let $P(x,y)$ be the assertion above Let any $t>0$ then $P\left(x,\sqrt{\frac{t}{f(x^2)}}\right)$ implies $f$ is strictly increasing, hence $f$ injective. $\textbf{Claim 1:}$ $f$ is not bounded on $\mathbb{R}^+$ Let $f(1)=V>0$, assume that for $x\in\mathbb{R}^+$ then $f(x)\le 1$. We have that $f$ is strictly increasing, then $$P(x,y)\implies f(xy)^2+f(x) \le 1\quad\forall x,y\in\mathbb{R}^+.$$Let's $y=1$ then $f(x)^2+f(x)\le 1\quad\forall y=x>0$ or $f(x)\le \sqrt{\frac{5}{4}}-\frac{1}{2}\quad\forall x>0.$ Similarly, we have $f(y)\le u_n\quad\forall y>0$, the sequence $(u_n)$ is defined by $u_1=1, u_{n+1}=\sqrt{u_n+\frac{1}{4}}-\frac{1}{2} \ \forall n\ge 1.$ It's easy to see that $(u_n)$ stricly decreasing and have limit is $0$, we have the contracdition. So there exist an $x_0>0$ such that $f(x_0)>1$. Then, by $P(1,x)$ we have $$f(1+x^2V)=V+f(x)^2\quad\forall x>0.$$Defined the sequence $(t_n)$ by $t_1=x_0$ and $t_{n+1}=1+t_n^2V$ then for $n\in\mathbb{N}$, we have $$f(t_{n+1})=V+f(t_n)^2>f(t_n)^2>\cdots>f(t_1)^{2^n}=f(x_0)^{2^n}.\quad\blacksquare$$$\textbf{Claim 2:}$ $\lim_{x\to 0^+} f(x)=0$ and $f$ is continous on $\mathbb{R}^+$. Because $f$ is strictly increasing so $\lim_{x\to 0^+}f(x)$ exists, assume it's $d$ with $d\ge 0$. If $d>0$ then there exist $d'>0$ such that $0<d'<d$ and $f(x)>d'\quad\forall x>0.$ By $P(x,y)$ and $f$ strictly increasing, we have $$f(d'y^2)<f(y^2f(x^2))<f(x+y^2f(x^2))=f(xy)^2+f(x)\quad\forall x,y>0.$$Let $x\to 0^+$ then $$f(d'y^2)\le d^2+d\quad\forall y>0.$$But this is asburd by $\textbf{Claim 1}$. Notice that give us $\lim_{x\to 0^+}f(x)=0$, now, for every $x>0$ we have $$\lim_{y\to 0^+}f(x+y^2f(x))=\lim_{y\to 0^+} (f(xy)^2+f(x))=f(x).\quad\blacksquare$$$\textbf{Claim 3:}$ The limit $\lim_{x\to 0^+}\frac{f(x)}{x}$ exist. By $\textbf{Claim 2}$ we have $f$ is continous or $\mathbb{R^+}$, $\lim_{x\to 0}f(x)=0$ and $f$ is unbound and so $f$ is surjective, let's $a>0$ such that $f(a)=1$. $$P\left(x,\frac{a}{x}\right)\implies f\left(x+a^2\frac{f(x^2)}{x^2}\right)=f(a)^2+f(x)=1+f(x)\quad\forall x>0.$$And so $$\lim_{x\to 0^+}f\left(x+a^2\frac{f(x^2)}{x^2}\right)=1=f(a).$$By the continuity of $f$ we have $\lim_{x\to 0^+}\left(x+a^2\frac{f(x^2)}{x^2}\right)=a$ or $$\lim_{x\to 0^+}\frac{f(x^2)}{x^2}=\frac{1}{a}.\quad \blacksquare$$$\textbf{Find all f.}$ $$P\left(x,\sqrt{\frac{t}{f(x^2)}}\right)\implies f(x+t)=f(x)+f\left(x\sqrt{\frac{t}{f(x^2)}}\right)^2\quad\forall x,t>0. \ (1)$$In $(1)$, for every $t$, let $x\to 0$ then $$f(t)=f\left(\sqrt{at}\right)^2\quad\forall t>0.$$Then $P(x,y)$ becomes $$P(x,y): f(x+y^2f(x^2))=f(x)+f\left(\frac{x^2y^2}{a}\right)\quad\forall x,y>0.$$$$P\left(x,\frac{\sqrt{ya}}{x}\right)\implies f\left(x+\frac{yaf(x^2)}{x^2}\right)=f(x)+f(y)=f\left(y+\frac{xaf(y^2)}{y^2}\right)\quad\forall x,y>0.$$By injectivity $$x+\frac{yaf(x^2)}{x^2}=y+\frac{xaf(y^2)}{y^2}\quad\forall x,y>0.$$Let $y=\sqrt{a}$ then $f(x^2)=\frac{x^2}{a}$ or $f(x)=\frac{x}{a}\quad\forall x>0$. Thus the solutions of $f$ is $$\boxed{f\equiv Vx, \ \forall x>0, \ \text{constant} \ V>0} \quad\blacksquare.$$

The answer is $f(x)=cx$ for some constant $c$, which clearly works. Clearly $f$ is strictly increasing by fixing $x$ and varying $y$. Now let $M=\inf f$ and suppose $M>0$. Fix some tiny positive real $\epsilon \ll M$ and find some $x,y$ such that $M \leq f(x)<f(x+y^2f(x^2)) \leq M+\varepsilon$. Then $f(xy)^2 \leq \varepsilon$, which is a contradiction. Thus $\inf f = 0$. In particular, this implies $\lim_{x \to 0} f(x)=0$, since $f$ is increasing. Thus by sending $y \to 0$ we find that $\lim_{y \to x^+} f(y)=f(x)$, hence $f$ is right continuous. To prove left-continuity, fix $x$ and let $x_1,x_2,\ldots$ be an increasing sequence of positive reals, all strictly less than $x$, with $\lim_{i \to \infty} x_i=x$. Let $y_1,y_2,\ldots$ be a sequence of positive reals such that $x_i+y_i^2f(x_i^2)=x$ for all $y$. Since $x_i,f(x_i^2)$ are increasing in $i$, $y_i$ must be decreasing, hence it has limit $0$. Then from $P(x_i,y_i)$ we find that $f(x)=f(x_iy_i)^2+f(x_i)$, and since $x_iy_i \leq xy_i$ which goes to $0$ as $i \to \infty$, it follows that $\lim_{i \to \infty} f(x_i)=f(x)$. This holds for any valid sequence of $x_i$, so $f$ is left continuous as well, hence continuous. Now consider $y=a/x$ and send $x \to 0$. Taking limits on both sides, we find that $$\lim_{x \to 0} f\left(x+a^2\frac{f(x^2)}{x^2}\right)=f(a)^2.$$Since $f$ is continuous, this means $\lim_{x \to 0} f(x^2)/x^2:=L$ exists. Furthermore, this limit is nonzero, since otherwise the above LHS limit is $\lim_{x \to 0} f(x)=0 \neq f(a)^2$. Thus we may evaluate this limit and write $f(a^2L)=f(a)^2$. This holds for all $a$, so the assertion can be rewritten as $$f(x+y^2f(x^2))=f(x^2y^2L)+f(x).$$Fix $x$. There should not exist any $y$ such that $x+y^2f(x^2)=x^2y^2L$, hence we require $f(x^2) \geq x^2L$. Since $f$ is increasing, this means $$f(x^2y^2L)+f(x)=f(x+y^2f(x^2)) \geq f(x+x^2y^2L),$$hence with a suitable substitution we have $f(a)+f(b) \geq f(a+b)$, i.e. $f$ is subadditive. In particular, this implies that $f(x) \leq nf(x/n)$ for positive integers $n$, which means $\tfrac{f(x)}{x} \leq \tfrac{f(x/n)}{x/n}$. Sending $n \to \infty$, it follows that $\tfrac{f(x)}{x} \leq L$, so $f(x) \leq Lx$ for all $x$. On the other hand, we also have $f(x) \geq Lx$, hence $f(x)=Lx$ for all $x$ and we obtain the desired solution set. $\blacksquare$

Cute! Claim 1: $f$ is strictly increasing and injective.

Let $f(1)=t$ then $f(t+1)=t(t+1)$ and $f(t^3+2t^2+t+1)=t^4+2t^3+t^2+t$ and therefore we have shown existence of $a,b$ such that $f(a)/a=f(b)/b$ $P(\sqrt{a},\sqrt{y/f(a)})-P(\sqrt{b},\sqrt{y/f(b)})$ gives $f(y+\sqrt{a})-f(y+\sqrt{b})=f(\sqrt{a})-f(\sqrt{b})$, so we have the well known FE $f(y+k)=f(y)+C$ for all $y > T$, now we get $f(\sqrt{k}\sqrt{x^2/f(x^2)})=C$ which implies $f \equiv cx$ for all $x \in \mathbb{R^{+}}$

Call the assertion $P(x,y)$. We claim that the only solutions are $f(x)=cx$ where $c$ is some positive constant. To show that this works, note \[cx+c^2x^2y^2=c\left(x+y^2\left(cx^2\right)\right)=f\left(x^2+y^2f\left(x^2\right)\right)=(cxy)^2+cx=f(xy)^2+f(x).\] We now prove a few claims. Claim: $f$ is strictly increasing. Proof. Suppose $x_2>x_1$ and let $d=x_2-x_1$. Then, $P\left(x_1,\frac d{f\left(x_1\right)^2}\right)$ gives $f\left(x_2\right)=f(x_1)+f(\text{something})^2>f(x_1)$ and hence the function is strictly increasing. $\blacksquare$ Claim: $\frac{f(x)}{x}$ is weakly increasing. Proof. Let $x_2>x_1$ be reals such that $\frac{f\left(x_1^2\right)}{x_1^2}<\frac{f\left(x_2^2\right)}{x_2^2}$. Let $f(x_1^2)=c_1x_1^2$ and $f(x_2^2)=c_2x_2^2$. Choose an arbitrary positive real $T$ and choose $y_1, y_2$ such that $T=x_1y_1=x_2y_2$. Then, comparing $P\left(x_1,y_1\right)$ and $P\left(x_2,y_2\right)$, we get \[f\left(x_1+y_1^2f\left(x_1^2\right)\right)=f\left(x_1y_1\right)^2+f\left(x_1\right)<f\left(x_2y_2\right)^2+f\left(x_2\right)=f\left(x_2+y_2^2f\left(x_2^2\right)\right)\]\[\Longrightarrow x_1+y_1^2c_1x_1^2<x_2+y_2^2c_2x_2^2\Longrightarrow x_2-x_1>(c_1-c_2)T^2\]However, $T$ can get arbitrarily large, so we get a contradiction. Hence, $c_1\leq c_2$. Claim: $\frac{f(x)}{x}$ is constant. Proof. Let $f(1)=c$. Note that $P(1,y)$ gives $f\left(1+cy^2\right)=f(y)^2+c$. So, if $\frac{f(y)}{y}=c$ then $\frac{f\left(1+cy^2\right)}{1+cy^2}=c$. Let $S$ be the set of all numbers $x$ such that $f(x)=cx$. Then, for any number $x\notin S$, there exist $x_1,x_2$ such that $x_1\leq x\leq x_2$. Since $\frac{f(x)}{x}$ is weakly increasing, we have \[c=\frac{f\left(x_1\right)}{x_1}\leq \frac{f(x)}{x}\leq \frac{f\left(x_2\right)}{x_2}=c\]which forces $f(x)=cx$, but this contradicts the fact that $x\notin S$. So, $S=\mathbb R^{+}$. This means that for all positive reals $x$, we have $f(x)=cx$, as desired. $\blacksquare$