Solve the system of equations $x^2=y^3, x^y=y^x$ in positive real numbers.
Taking $\ln$ of the second equation, we have $\ln x/x=\ln y/y$. This function is increasing in $(0,e)$ and decreasing in $(e,\infty)$. For any value of $x\neq e$, we can find a unique value of $y$ such that $x^y=y^x$. But how can we find a closed form to substitute into $x^2=y^3$?
On
Write $y=x^{\frac{2}{3}}$, then
$$x^{x^{\frac 23}}=x^{\frac{2x}{3}}$$
so
$$x^{\frac{2}{3}}=\frac{2x}{3}$$
Can you continue?
On
Notice its better to start with the expressing y in terms of x because here you have $x^2 = y^3$ if you start expressing $x$ in terms of $y$ you would have to take the square root which would involve +- so you can start something like this $x^2 = y^3 if y = x^{2/3}. $ now substitute that into the other formula and after few algebric manipulation you will get the answer.
Why don't you continue with the logarithms ? $$\frac{\ln x}x=\frac{\ln y}y,\\\color{blue}{2\ln x=3\ln y},$$ then dividing memberwise, $$2x=3y,$$ or $$\color{blue}{\ln x+\ln 2=\ln y+\ln3}.$$
You now have two linear equations in $\ln x$ and $\ln y$.