$$\frac{\cos \left(3(x - \frac{\pi}{4})\right)}{\cos(x - \frac{\pi}{4})} = \frac{\sin 3x - \cos 3x}{\sin x + \cos x}$$
My attempt: \begin{align} LHS &=\frac{\cos \left(3(x - \frac{\pi}{4})\right)}{\cos(x - \frac{\pi}{4})} \\ &= 4\cos^2 (x - \frac{\pi}{4}) - 3 \\ &= 4(\cos x \cos\frac \pi 4 + \sin \frac \pi 4\sin x )^2 - 3 \\ &=-1 + 4\sin x \cos x \\ RHS &= \frac{\sin 3x - \cos 3x}{\sin x + \cos x} \\ &= \frac{2\sin x - 4\sin^3x - 4\cos^3x + 3\cos x}{\sin x + \cos x} \\ &= -1 + 4\sin x\cos x \end{align}
But it seems that there is a way to show it more efficiently, but I am not sure how.
Using the angle addition formula $$\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$$ we have
$$\frac{\cos(3(x-\frac{\pi}{4}))}{\cos(x-\frac{\pi}{4})}=\frac{\cos(3x)\cos(\frac{3\pi}{4})+\sin(3x)\sin(\frac{3\pi}{4})}{\cos(x)\cos(\frac{\pi}{4})+\sin(x)\sin(\frac{\pi}{4})}$$ $$=\frac{-\frac{\sqrt{2}}{2}\cos(3x)+\frac{\sqrt{2}}{2}\sin(3x)}{\frac{\sqrt{2}}{2}\cos(x)+\frac{\sqrt{2}}{2}\sin(x)}$$ $$=\frac{\sin(3x)-\cos(3x)}{\cos(x)+\sin(x)}$$