$$\frac{a^3 + b^3}{a^2 + b^2} = (a + b) \frac{a^2 - ab + b^2}{a^2 + b^2} = 1 - \frac{ab}{a^2 + b^2}.$$
From this the right inequality is evident with equality for $ab = 0$, i.e. for $a = 0$, $b = 1$ and for $a = 1$, $b = 0$.

The left inequality is equivalent to
$$\frac{1}{2} \le 1 - \frac{ab}{a^2 + b^2} \iff \frac{ab}{a^2 + b^2} \le \frac{1}{2} \iff 2ab \le a^2 + b^2 \iff 0 \le (a - b)^2.$$
This inequality is obvious with equality for $a = \frac{1}{2}$. In this case also $b = \frac{1}{2}$.