$\mathrm{y}={\left(\mathrm{sinx}\right)}^{{\left(\mathrm{sinx}\right)}^{\left(\mathrm{sinx}\right)...\infty \mathrm{times}}}$ then $\frac{\mathrm{dy}}{\mathrm{dx}}=$

$y={\left(\sin x\right)}^y$

Apply logarithm on both sides

$\log y=y\log \left(\sin x\right)$

Differentiate both sides

$\begin{array}{rcl}\frac{1}{y}\left(\frac{dy}{dx}\right)& =& y\left(\frac{1}{\sin x}\cos x\right)+\log \sin x\left(\frac{dy}{dx}\right)\\ & =& ycotx+\mathrm{logsin}x\left(\frac{dy}{dx}\right)\\ \frac{dy}{dx}\left(\frac{1}{y}-\log \sin x\right)& =& y\left(cotx\right)\\ \frac{dy}{dx}& =& \boxed{\frac{{\mathbf{y}}^{\mathbf{2}}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{x}}{\mathbf{1}\mathbf{-}\mathbf{y}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}}\end{array}$