If y = log (sec e^x^2), then dy/dx=

$y = \log (\sec e^{x^2})$

so $\frac{dy}{dx} = \frac{1}{\sec e^{x^2}} \frac{d}{dx}(\sec e^{x^2})$

$= \frac{\sec e^{x^2} \tan e^{x^2}}{\sec e^{x^2}} \frac{d}{dx} (e^{x^2})$

$\frac{dy}{dx} = 2x e^{x^2} \tan e^{x^2}$