, 2015

Answer: $f(z) = 0$ or $f(z) = z^2$.
Replacing $y$ by $-y$ gives us
$$f(x) + f(y)^2 + \frac{2f(xy)}{x} = f(x + y^2) = f(x) + f(-y)^2 + \frac{2f(-xy)}{x},$$
which implies that
$$f(y)^2 + \frac{2f(xy)}{x} = f(-y)^2 + \frac{2f(-xy)}{x} \quad (7)$$
for all $x \neq 0$ and all $y$. Let $x = 1$ and complete the squares:
$$(f(y) + 1)^2 = f(y)^2 + 2f(y) + 1 = f(-y)^2 + 2f(-y) + 1 = (f(-y) + 1)^2.$$
Hence
$$f(y) + 1 = \pm (f(-y) + 1),$$
so that, for any $y$,
either $f(y) = f(-y)$ or $f(y) + f(-y) = -2$.
Suppose $f(y) + f(-y) = -2$ for all $y \neq 0$. Equation (7) then simplifies to
$$f(xy) + 1 = x(f(y) + 1), \quad x, y \neq 0.$$

Let $y = 1$ (or swap $x$ and $y$) to deduce
$$f(x) = (1 + f(1))x - 1 = ax - 1, \quad x \neq 0,$$
where $a = 1 + f(1)$. Insert this expression into the original equation:
$$(a^2 - a)y^2 + 1 = \frac{2}{x}, \quad x, y, x + y^2 \neq 0.$$
This is clearly impossible (fix one $y \neq 0$ and let $x = 1$ and $x = 2$).

The contradiction shows that $f(-y) = f(y)$ for some $y \neq 0$. Then $f(-xy) = f(xy)$ for all $x \neq 0$ by (7), so that $f(-z) = f(z)$ for all $z$.
Returning to the original equation, let $y = 1$:
$$f(x+1) = \left(1 + \frac{2}{x}\right)f(x) + f(1)^2, \quad x \neq 0.$$
Exchange $x$ for $-x - 1$ and use $f(-x) = f(x)$:
$$f(x) = \left(1 - \frac{2}{x+1}\right)f(x+1) + f(1)^2, \quad x \neq -1.$$
Eliminate $f(x+1)$ from these two equations and simplify:
$$f(x) = x^2 f(1)^2, \quad x \neq 0, -1. \qquad (8)$$
This formula is, in fact, accurate even for $x = 0$ and $x = -1$. Indeed, letting $y = 0$ in the original equation leads to
$$f(0) = 0 = 0^2 f(1)^2.$$
Moreover, letting $x = 1$ in (8) yields $f(1) = f(1)^2$, so that
$$f(-1) = f(1) = (-1)^2 f(1)^2.$$
Finally, $f(1) = f(1)^2$ determines $f(1) = 0$ or $f(1) = 1$. Therefore, (8) provides two easily verified solutions $f(z) = 0$ and $f(z) = z^2$. $\square$