$\text{ The function }y=f\left(x\right)\text{ whose graph passes through }(0,0)\text{ and whose derivative is }$$\frac{{\cos }^{4}x+{\sin }^{4}x}{{\cos }^{2}x-{\sin }^{2}x}\text{ is given by }$

$\begin{array}{l}f\left(x\right)=\int \frac{{\cos }^{4}x+{\sin }^{4}x}{{\cos }^{2}x-{\sin }^{2}x}dx\\ \text{ Where result }2\left({\cos }^{4}x+{\sin }^{4}x\right)\\ ={\left({\cos }^{2}x+{\sin }^{2}x\right)}^2+{\left({\cos }^{2}x-{\sin }^{2}x\right)}^2\\ =1+{\left({\cos }^{2}x-{\sin }^{2}x\right)}^2\\ \therefore f\left(x\right)=\frac{1}{2}\int \frac{{\left({\cos }^{2}x-{\sin }^{2}x\right)}^2+1}{\left({\cos }^{2}x-{\sin }^{2}x\right)}dx\\ =\frac{1}{2}\int \left[\left({\cos }^{2}x-{\sin }^{2}x\right)+\frac{1}{{\cos }^{2}x-{\sin }^{2}x}\right]dx\end{array}$

$\begin{array}{l}=\frac{1}{2}\left[\int \cos 2xdx+\int \sec 2xdx\right]\\ =\frac{1}{2}\left[\frac{\sin 2x}{2}+\frac{1}{2}\log \left[\tan \left[\frac{2x}{2}+\frac{\pi }{4}\right]\right]+C\right.\\ f\left(x\right)=\frac{1}{4}\sin 2x+\frac{1}{4}\log \left[\tan \left(x+\frac{\pi }{4}\right)\right]+C\\ \text{ Given the curve }f\left(x\right)\text{ pass through }(0,0)\\ \therefore f\left(o\right)=0\\ \Rightarrow 0=0+0+C\\ \Rightarrow C=0\\ \text{ Hence, }f\left(x\right)=\frac{1}{4}\sin 2x+\frac{1}{4}\log \left|\tan \left(x+\frac{\pi }{4}\right)\right|\end{array}$