Since we have $x^2 + y^2 = 1$, it definitely follows that $|x| \le 1$ and $|y| \le 1$ hold. Defining a function $g_k(x) := x - x^{2k+1}$, the signs of $x$ and $g_k(x)$ are therefore equal, and we have $g_k(-x) = -g_k(x)$. The given function can be expressed as $f_k(x, y) = g_k(x) + g_k(y)$, and we certainly have $f_k(x, y) \le f_k(|x|, |y|)$. Since $x^2 + y^2 = 1$ implies $|x|^2 + |y|^2 = 1$, we can assume that $x, y \ge 0$ holds, which allows us to apply the means inequality. For the quadratic mean, we have
$$m_2(x, y) = \sqrt{\frac{x^2 + y^2}{2}} = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2},$$
and from
$$x + y = 2m_1(x, y) \le 2m_2(x, y) \le 2m_{2k+1}(x, y)$$
we obtain
$$-(x^{2k+1} + y^{2k+1}) = -2m_{2k+1}^2(x, y) \le -2m_2^2(x, y).$$
We therefore have both $x+y \le \sqrt{2}$ and
$$-(x^{2k+1} + y^{2k+1}) \le \frac{2}{\sqrt{2}^{2k+1}} = \frac{\sqrt{2}}{2^k},$$
which imply
$$f_k(x, y) \le \frac{2^k - 1}{2^k} \sqrt{2}$$
with equality holding for $x = y = \frac{\sqrt{2}}{2}$.
qed