Without loss of generality we may assume that $a, b, c, d \ge 0$. By dividing $2$, it is enough to prove the inequality for the numbers $0 \le a, b, c, d \le 1$:
$$\frac{a^3}{b^2+1} + \frac{b^3}{c^2+1} + \frac{c^3}{d^2+1} + \frac{d^3}{a^2+1} \le 2.$$
Now, suppose that $a$ is the largest among the $a, b, c, d$. Then,
$$\frac{a^3}{b^2+1} + \frac{b^3}{c^2+1} \le \frac{1}{2} + \frac{a^3}{c^2+1}.$$
Indeed, clearing denominators and simplifying gives
$$2a^3b^2 + b^2c^2 + b^2 + c^2 + 1 \ge 2a^3c^2 + 2b^5 + 2b^3$$
This follows from the inequality $b^2c^2+1 \ge b^2+c^2$. $b^2(a^3-b^3)+b^2(1-b)+c^2(1-a^3) \ge 0$
Similarly,
$$\frac{a^3}{c^2+1} + \frac{c^3}{d^2+1} \le \frac{1}{2} + \frac{a^3}{d^2+1} \quad \text{and} \quad \frac{a^3}{d^2+1} + \frac{d^3}{a^2+1} \le \frac{1}{2} + \frac{a^3}{a^2+1}.$$
Clearly, $2a^3 \le a^2 + 1$, so the problem is solved.