It is deduced, from the state of the problem that
$$(x^2 + y^2)(x + y) = 1$$
Without loss of generality, assume that $x > y$. Therefore we have
$$\begin{align*} \frac{x-y}{x^6-y^6} &\le \frac{4}{3}(x+y) \\ \Leftrightarrow \quad &x^6 - y^6 \ge \frac{3}{4}(x-y) \\ \Leftrightarrow \quad &x^6 - y^6 \ge \frac{3}{4}(x^2+y^2)(x-y) \\ \Leftrightarrow \quad &x^6 - y^6 \ge \frac{3}{4}(x^2+y^2)(x^4-y^4) \\ \Leftrightarrow \quad &x^6 - y^6 \ge -3x^2y^4 + 3x^4y^2 \\ \Leftrightarrow \quad &x^6 - y^6 \ge 3x^2y^2(x^2-y^2) \\ \Leftrightarrow \quad &(x^2-y^2)(x^4-2x^2y^2+y^4) \ge 0 \\ \Leftrightarrow \quad &(x^2-y^2)^3 \ge 0 \\ \Leftrightarrow \quad &x \ge y \end{align*}$$
Which is true. Hence, the inequality holds.