Given the function $y = (\cos x)^{x^2}$, we need to find its derivative $\frac{dy}{dx}$.

To solve this, we will use logarithmic differentiation. Let's start by taking the natural logarithm of both sides:

$\ln y = \ln \left( (\cos x)^{x^2} \right)$

Using the logarithm power rule $\ln(a^b) = b \ln a$, we get:

$\ln y = x^2 \ln (\cos x)$

Next, differentiate both sides with respect to $x$:

$\frac{d}{dx} (\ln y) = \frac{d}{dx} \left( x^2 \ln (\cos x) \right)$

On the left side, using the chain rule:

$\frac{d}{dx} (\ln y) = \frac{1}{y} \frac{dy}{dx}$

On the right side, using the product rule $\frac{d}{dx} (u \cdot v) = u'v + uv'$, where $u = x^2$ and $v = \ln (\cos x)$:

$\frac{d}{dx} \left( x^2 \ln (\cos x) \right) = 2x \ln (\cos x) + x^2 \cdot \frac{d}{dx} (\ln (\cos x))$

To differentiate $\ln (\cos x)$:

$\frac{d}{dx} (\ln (\cos x)) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x$

Therefore:

$\frac{d}{dx} \left( x^2 \ln (\cos x) \right) = 2x \ln (\cos x) + x^2 (-\tan x)$

Combining everything:

$\frac{1}{y} \frac{dy}{dx} = 2x \ln (\cos x) - x^2 \tan x$

Multiplying both sides by $y$:

$\frac{dy}{dx} = y \left( 2x \ln (\cos x) - x^2 \tan x \right)$

Recall that $y = (\cos x)^{x^2}$:

$\frac{dy}{dx} = (\cos x)^{x^2} \left( 2x \ln (\cos x) - x^2 \tan x \right)$

Rewriting the expression:

$\frac{dy}{dx} = (\cos x)^{x^2} \left( 2x \ln (\cos x) - x^2 \tan x \right)$

Comparing with the given options, the correct answer is:

a. ((\cos x)^{x^2} (-