There are...Are they?

This equation can be written as $y \,=\, x^y$ , and so $x \,=\, y^{1/y}$ . The graph of $y = f(x) = x^{1/x}$ is increasing for $0 \le x \le e$ , rising from $y=0$ to $y = e^{1/e}$ , and then decreasing for $x > e$ , tending to a limit of $1$ . This function is injective for $0 \le x \le e$ (or, at least, that is the domain for which it is injective with largest range). We are interested in the inverse function $f^{-1}$ , which has domain $0 \le x \le e^{1/e}$ and range $0 \le y \le e$ . Since $2^{1/2} = \sqrt{2}$ and $0 \le 2 \le e$ , we deduce that $f^{-1}(\sqrt{2}) = 2$ . The fact that $4^{1/4} = \sqrt{2}$ is irrelevant (unless we choose to choose a different domain for $f$ ).

The answer is $2$ .

$y(\sqrt{2})=2$

$y(\sqrt{2})= {\sqrt{2}}^{2}$

$\large y(\sqrt{2})= {\sqrt{2}}^{{\sqrt{2}}^{2}}$

$\large y(\sqrt{2})= {\sqrt{2}}^{{\sqrt{2}}^{{\sqrt{2}^{.^{.^{.}}}}}}$

So, $y(\sqrt{2})=\boxed{2}$