<s>[INST] Find $y$ for $y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$ where $y$ is a positive real number Here is the answer in the textbook with which I disagree / don't understand:
$$y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$$
$$y^{y^{\frac{3}{2}}} = (y^{\frac{3}{2}})^y = y^{\frac{3}{2}y}$$
Comparing indices, we get:
$$y^{\frac{3}{2}} = \frac{3}{2}y$$
from $y^{\frac{1}{2}} = {\frac{3}{2}}$, we get: $$y = {\frac{9}{4}}$$
Here what I came up with:
$$y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y$$
$$(\sqrt[y]{y})^y = (y^{\frac{1}{y}})^y = y$$
$$y^{\sqrt[y]{y}} = y \Rightarrow \sqrt[y]{y} = 1$$
And I get stuck somewhere around: $y = 1^y$
 [/INST] For positive $y$ you have  $(\sqrt[y]{y})^y = y$  
so $y^{\sqrt[y]{y}} = (\sqrt[y]{y})^y\implies y^{\sqrt[y]{y}} = y \implies y^{\sqrt[y]{y}-1} = 1$
so either $y=1$ or $\sqrt[y]{y}-1=0$
in both cases implying $y=1$

Alternatively if the original question is actually $y^{y \sqrt{y}} = (y \sqrt{y})^y$
Then this indeed gives $y^{y^{3/2}}=y^{3y/2}$
so either $y=1$ or $y^{3/2}=\frac{3y}{2}$
i.e. either $y=1$ or $y^{1/2}=\frac{3}{2}$
i.e. either $y=1$ or $y=\frac{9}{4}$
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