Differentiate $f ( x ) = \frac { \ln \left( x ^ { 2 } \cos ( x ) \right) } { \sqrt { 1 - x ^ { 2 } } }$ could I get a clue as I dont even know where to begin , I thought about the quotient rule but thats making another quotient rule necessary and I have a cos x within a ln , I really dont even know where to begin for me to show my own working of the problem.

I would use logarithmic differentiation. Setting $u(x)=x^2\cos x$, we have
\begin{align}
\frac{f'(x)}{f(x)}&=\frac{u'(x)}{u(x)\ln u(x)}-\frac12\frac{-2x}{1-x^2} =\frac{2x\cos x-x^2\sin x}{x^2\cos x\,\ln(x^2\cos x)}-\frac12\frac{-2x}{1-x^2}\\[1ex]
&=\frac{2\cos x-x\sin x}{x\cos x\,\ln(x^2\cos x)}+\frac x{1-x^2},
\end{align}
whence
$$f'(x)=\frac{f'(x)}{f(x)}\,f(x)=\frac{2\cos x-x\sin x}{x\cos x\sqrt{1-x^2}}+\frac{x\ln(x^2\cos x)}{(1-x^2)^{3/2}}.$$
