Answer
$y^2=\sin^{-1} (2 \tan x+c) $
Work Step by Step
Given: $y \dfrac{dy}{dx}=\sec y^2\sec^2 x$
Separate the differentials and integrate.
$\int \sec^2 xdx=\int y \cos y^2 dy$
or, $\int \sec^2 xdx=(\dfrac{1}{2} \int \cos y^2 (2y) dy$
This implies that $(\dfrac{1}{2}) \sin y^2=\tan x+c$
Thus, $\sin^{1} (\sin y^2)=\sin^{-1} (2 \tan x+c) \implies y^2=\sin^{-1} (2 \tan x+c) $