Answer
$$ y''= 2\sec x^2+4x^2 \sec x^2 \tan x^2.$$
Work Step by Step
Recall that $(\tan x)'=\sec^2 x$.
Recall that $(x^n)'=nx^{n-1}$
Given $ y=\tan x^2$, then, by using the chain rule, we have
$$ y'=(2x)\sec x^2=2x\sec x^2,$$
and, by using the product rule, the second derivative $ y''$ is given by
$$ y''=2\sec x^2+2x (2x) \sec x^2 \tan x^2 \\=2\sec x^2+4x^2 \sec x^2 \tan x^2.$$
