Solve for $y$ in the given equation Solve:
$$y^{y\sqrt {y}}=(y\sqrt {y})^{y}$$

My Attempt:
$$y^{y\sqrt {y}}=(y\sqrt {y})^{y}$$
$$y^{y^{\dfrac {3}{2}}}=(y^{\dfrac {3}{2}})^y$$

How do I solve further?
The domain is $y&gt;0$.

We need to solve 
$$y^{y\sqrt{y}}=y^{\frac{3}{2}y},$$
which gives $y=1$ or
$$y\sqrt{y}=\frac{3}{2}y,$$
which is $y=\frac{9}{4}$ and we get the answer: $\{1,\frac{9}{4}\}$.
$y^{y^3\over 2}={y^y} {y^{y\over 2}}$

$y^{y^3\over 2}={y^3y\over 2}$

$\frac{y^3} 2 =\frac {3y} 2 $

$2{y^3\over 2}=3y$

$4{y^3}=9y^2$

$4{y^3}-9{y^2}=0$

${y^2}(4y-9)=0$

$SS={\{0,0,{9\over 4}}\}$

However if you define $0^0$ as an indeterminate form the only root to this equation is $9\over 4$.