All functions $\frac{1}{f\left(y^2f(x)\right)} = \big(f(x)\big)^2\left(\frac{1}{f\left(x^2-y^2\right)} + \frac{2x^2}{f(y)}\right)$ How can I find all continuous functions $f:\mathbb{R} \rightarrow \mathbb{R}^+$ such that
$$\frac{1}{f\left(y^2f(x)\right)} = \big(f(x)\big)^2\left(\frac{1}{f\left(x^2-y^2\right)} + \frac{2x^2}{f(y)}\right)$$
for all reals $x,y$?
 A: Let's first show that $f(x) = f(-x)$ for all $x \in \mathbb{R}$. Put $x=y=0$, and we get
$$\frac{1}{f(0)} = f(0),$$
which implies that $f(0) = 1$. Now letting $x=0$ we get
$$\frac{1}{f(y^2)} = \frac{1}{f(-y^2)},$$
which implies that $f(y^2) = f(-y^2)$, proving our first claim.
Now since $f(x) = f(-x)$ there exists a function $g\colon [0,\infty) \to \mathbb{R}^+$ such that $f(x) = g(x^2)$. The condition on $f$ becomes
$$\frac{1}{g(y^4g^2(x^2))} = (g(x^2))^2 \left( \frac{1}{g((x^2 - y^2)^2)} + \frac{2x^2}{g(y^2)} \right).$$
By changing $x^2 \mapsto x$ and $y^2 \mapsto y$, this turns to
$$\frac{1}{g(y^2 g^2(x))} = (g(x))^2 \left( \frac{1}{g((x-y)^2)} + \frac{2x}{g(y)}\right).$$
Now notice that we already know that $g(0) = f(0) = 1$.
Also, by setting $x=1$, $y=0$ we get
$$1 = (g(1))^2 \left( \frac{1}{g(1)} + 2 \right) \Leftrightarrow g(1) = (g(1))^2 + 2 (g(1))^3 \Leftrightarrow$$
$$1 = g(1) + 2 (g(1))^2 \Leftrightarrow 2(g(1) - \frac{1}{2})(g(1) + 1) = 0.$$
Hence $g(1) = \frac{1}{2}$.
We will show by induction that $g(n) = \frac{1}{1+n}$ for all $n \in \mathbb{N}$. Suppose that the claim holds for some $n \in \mathbb{N}$, then
$$\frac{1}{g((n+1)^2 g^2(n))} = (g(n))^2 \left( \frac{1}{g((n-(n+1))^2)} + \frac{2n}{g(n+1)}\right) \Leftrightarrow$$
$$\frac{1}{g(1)} = \frac{1}{(n+1)^2} \left( \frac{1}{g(1)} + \frac{2n}{g(n+1)}\right) \Leftrightarrow$$
$$2 (n+1)^2 = 2 + \frac{2n}{g(n+1)} \Leftrightarrow g(n+1) = \frac{1}{n+2},$$
and by induction we have that
$$g(n) = \frac{1}{n+1}$$
for all $n \in \mathbb{N}$.
Consider now the original condition on $g$ and let $x$ and $y$ be natural numbers. We get that
$$\frac{1}{g(\frac{y^2}{(x+1)^2})} = \frac{1}{(x+1)^2} \left( (x-y)^2 + 1 + 2x(y+1) \right) \Leftrightarrow$$
$$\frac{1}{g(\frac{y^2}{(x+1)^2})} = \frac{1}{(x+1)^2} \left( (x+1)^2 + y^2 \right) \Leftrightarrow$$
$$g(\frac{y^2}{(x+1)^2}) = \frac{(x+1)^2}{(x+1)^2 + y^2} = \frac{1}{\frac{y^2}{(x+1)^2} + 1},$$
and hence the formula
$$g(x) = \frac{1}{x+1}$$
holds for all squares of rational numbers. But they are dense in $[0,\infty)$ and since $g$ was continuous, we get that the only solution is
$$g(x) = \frac{1}{x+1}.$$
(Checking that this is indeed a solution is straightforward.)
Thus
$$f(x) = \frac{1}{x^2 + 1}.$$
