Determine the first derivative $(dy/dx)of\text{}y=\mathrm{tan}(3{x}^{2})$

Let $u=3{x}^{2}$, then

$y=\mathrm{tan}\left(u\right)$

And, $\frac{dy}{dx}=(\frac{dy}{du})\cdot (\frac{du}{dx})$

Taking $y=\mathrm{tan}\left(u\right)$, then $\frac{dy}{du}={\mathrm{sec}}^{2}\left(u\right)$

Taking $u=3{x}^{2}$, then $\frac{du}{dx}=6x$

So, $\frac{dy}{dx}=(\frac{dy}{du})\cdot (\frac{du}{dx})$

$\frac{dy}{dx}=\left({\mathrm{sec}}^{2}\left(u\right)\right)\cdot \left(6x\right)$

$\frac{dy}{dx}=6x\cdot {\mathrm{sec}}^{2}\left(3{x}^{2}\right)$