Let $P$ denote the assertion that:
$$f(x^3 + x f(xy)) = f(xy) + x^2 f(x + y)$$
By $P(x, \frac{y}{x})$ we have
$$A(x, y) : \quad f(x^3 + x f(y)) = f(y) + x^2 f\left(x + \frac{y}{x}\right)$$
If there exists a pair $(x, y)$ such that $f(x^3 + x f(y)) = 0$. Then
$$A(x, y) \rightarrow f(y) + x^2 f\left(x + \frac{y}{x}\right) = 0 \quad (1)$$
From the definition of function we know that $\forall x \ge 0 : f(x) \ge 0$ so (1) infers that $f(y) = 0$. Thus it suffices to prove for each $t \ge 0$ there exists $x \ge 0$ such that $f(x^3 + x f(t)) = 0$.

Consider the polynomial $P(x) = x^3 + x f(1) - 1$. It has at least one positive real root, namely $x_0$ since $P(0) < 0$ and the leading coefficient of polynomial is positive. From $A(x_0, 1)$ we have
$$f(x_0^3 + x_0 f(1)) = f(1) \implies f\left(x_0 + \frac{1}{x_0}\right) = 0$$
Thus there is a positive number $c$ such that $f(c) = 0$. For any non-negative number $t$ consider the polynomial $Q_t(x) = x^3 + x f(t) - c$. Again $Q_t$ has at least one positive root like $x_1$ because $Q_t(0) < 0$ and the leading coefficient is positive. Then
$$f(x_1^3 + x_1 f(t)) = f(c) = 0$$
And $x_1$ is what we supposed to find for each $t$ to conclude $f(t) = 0$. ■