If $y=e^{x^2+e^{x^2+e^{x^2}+\cdots }}$ then $\frac{dy}{dx}=$

We have to differentiate 

$y=e^{x^2+e^{x^2+e^{x^2}+\cdots }}$

Reducing,  $y=e^{x^2+e^{x^2+e^{x^2}+\cdots }}=e^{x^2+y}$

So, we have an implicit function now,

$y=e^{x^2+y}$
Differentiating both the sides w.r.t $x$,

$\frac{dy}{dx}=\frac{d\left(e^{x^2+y}\right)}{dx} \Rightarrow \frac{dy}{dx}=e^{x^2+y}\times \frac{d\left(x^2+y\right)}{dx}\Rightarrow \frac{dy}{dx}=e^{x^2+y}\times \left(2x+\frac{dy}{dx}\right)$

$\frac{dy}{dx}=y\left(2x+\frac{dy}{dx}\right)\Rightarrow \frac{dy}{dx}-y\frac{dy}{dx}=2xy \Rightarrow \frac{dy}{dx}=\frac{2xy}{1-y}$