, 2020

$$2f(y^2) = f(y)^2. \qquad (4)$$
Using this, we can cancel the $2f(y^2)$ on the left hand side of the original functional equation against the $f(y)^2$ on the right hand side:
$$f(x^2y) = x^2f(y).$$
Substituting $y = 1$ in this equation yields $f(x^2) = x^2f(1)$, and substituting $y = -1$ yields $f(-x^2) = x^2f(-1)$. Because $x^2$ takes on all non-negative numbers as value when $x \in \mathbb{R}$, we get
$$f(x) = \begin{cases} cx & \text{als } x \ge 0, \\ dx & \text{als } x < 0, \end{cases}$$
with $c = f(1)$ and $d = -f(-1)$. Now substitute $y = 1$ in equation (4), which yields $2f(1) = f(1)^2$, hence $2c = c^2$. It follows that $c = 0$ or $c = 2$. If we actually substitute $y = -1$ in equation (4), then we find that $2f(1) = f(-1)^2$, hence $2c = (-d)^2$. For $c = 0$, we get $d = 0$, and for $c = 2$, we get $d = 2$ or $d = -2$. So, there are three cases:
* $c = 0, d = 0$: then $f(x) = 0$ for all $x$;

* $c = 2, d = 2$: then $f(x) = 2x$ for all $x$;
* $c = 2, d = -2$: then $f(x) = 2x$ for $x \ge 0$, and $f(x) = -2x$ for $x < 0$, or, in other words, $f(x) = 2|x|$ for all $x$.
Using the first function, both sides of the functional equation become 0, so this function is a solution. Using the second function, we get $2x^2y + 4y^2$ on the left hand side, and $(x^2 + 2y) \cdot 2y = 2x^2y + 4y^2$ on the right hand side, so this function is a solution as well. Using the third function, we get $2|x^2y| + 4|y^2| = 2x^2|y| + 4y^2$ on the left hand side, and $(x^2 + 2|y|) \cdot 2|y| = 2x^2|y| + 4|y|^2 = 2x^2|y| + 4y^2$ on the right hand side, so also this function is a solution.
Altogether, we found the three solutions: $f(x) = 0$, $f(x) = 2x$, and $f(x) = 2|x|$. $\square$