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Given \(\tan^{-1}(x^{2}-y^{2})=a\), we differentiate both sides with respect to $x$.
Using the chain rule, we have: $$\frac{d}{dx} \tan^{-1}(x^{2}-y^{2}) = \frac{d}{dx} (a)$$ $$\frac{1}{1+(x^{2}-y^{2})^{2}} \cdot \frac{d}{dx}(x^{2}-y^{2}) = 0$$
Now, differentiate $x^{2}-y^{2}$ with respect to $x$: $$\frac{d}{dx}(x^{2}-y^{2}) = 2x - 2y\frac{dy}{dx}$$
Substitute this back into the equation: $$\frac{1}{1+(x^{2}-y^{2})^{2}} \cdot (2x - 2y\frac{dy}{dx}) = 0$$
Since $\frac{1}{1+(x^{2}-y^{2})^{2}}$ cannot be zero, we must have: $$2x - 2y\frac{dy}{dx} = 0$$ $$2y\frac{dy}{dx} = 2x$$ $$\frac{dy}{dx} = \frac{2x}{2y}$$ $$\frac{dy}{dx} = \frac{x}{y}$$
Final Answer: \(\frac{x}{y}\)