How do you differentiate tan^2(3-x^2)?

$y ' = 2 \tan \left(3 - {x}^{2}\right) \cdot \frac{1}{\cos} ^ 2 \left(3 - {x}^{2}\right) \cdot \left(- 2 x\right) =$
$= - 4 x \cdot \tan \left(3 - {x}^{2}\right) \cdot \frac{1}{\cos} ^ 2 \left(3 - {x}^{2}\right)$.