Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ for which

$$
f\left(x^{2}\right)-f\left(y^{2}\right) \leq(f(x)+y)(x-f(y))
$$

for all $x, y \in \mathbb{R}$.
Substituting $x=y=0$ gives $0 \leq f(0) \cdot -f(0)$. But squares are nonnegative, so it follows that $f(0)^{2}=0$ and thus $f(0)=0$. Now, substituting $x=0$ and $y=t$ gives $-f\left(t^{2}\right) \leq t \cdot -f(t)$, while substituting $x=t$ and $y=0$ gives $f\left(t^{2}\right) \leq f(t) \cdot t$. We find $t f(t) \leq f\left(t^{2}\right) \leq t f(t)$, so $f\left(t^{2}\right)=t f(t)$ for all $t \in \mathbb{R}$. We can now replace $f\left(x^{2}\right)-f\left(y^{2}\right)$ on the left side of the functional inequality with $x f(x)-y f(y)$. Expanding the right side, we get $x f(x)-y f(y)+x y-f(x) f(y)$, so we have

$$
f(x) f(y) \leq x y
$$

From $f\left(t^{2}\right)=t f(t)$, it follows that $f(1)=-f(-1)$. By first taking $y=1$ and then $y=-1$, we get for all $x \in \mathbb{R}$:

$$
x \geq f(x) f(1)=-f(x) f(-1) \geq -x \cdot -1=x
$$

Thus, equality must always hold, which means that $f(x) f(1)=x$. This particularly implies $f(1)^{2}=1$, so $f(1)=1$ or $f(1)=-1$. In the first case, we get $f(x)=x$ for all $x$, and in the second case, $f(x)=-x$ for all $x$.
If we check $f(x)=x$ in the original functional inequality, we get $x^{2}-y^{2}$ on the left and $(x+y)(x-y)=x^{2}-y^{2}$ on the right, so this function satisfies the inequality. If we check $f(x)=-x$, we get $-x^{2}+y^{2}$ on the left and $(-x+y)(x+y)=-x^{2}+y^{2}$ on the right, so this function also satisfies the inequality. We conclude that the solutions are: $f(x)=x$ for all $x \in \mathbb{R}$ and $f(x)=-x$ for all $x \in \mathbb{R}$.