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we have,

$f(x)=\sin x-\cos x$

$f^{\prime}(x)=\cos x+\sin x$

$=\sqrt{2}\left(\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right)$

$=\sqrt{2}\left(\frac{\sin \pi}{4} \cos x+\frac{\cos \pi}{4} \sin x\right)$

$=\sqrt{2} \sin \left(\frac{\pi}{4}+x\right)$

Now,

$x \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$

$\Rightarrow-\frac{\pi}{4}

$\Rightarrow 0<\frac{\pi}{4}+x<\frac{\pi}{2}$

$\Rightarrow \sin 0^{\circ}<\sin \left(\frac{\pi}{4}+x\right)<\sin \frac{\pi}{2}$

$\Rightarrow 0<\sin \left(\frac{\pi}{4}+x\right)<1$

$\Rightarrow \sqrt{2} \sin \left(\frac{\pi}{4}+x\right)>0$

$\Rightarrow f^{\prime}(x)>0$

Hence, $f(x)$ is an increasing function on $(-\pi / 4, \pi / 4)$