Answer:
\[y=b{{\tan }^{-1}}\left( \frac{x}{a}+{{\tan }^{-1}}\frac{y}{x} \right)\] \[\Rightarrow \] \[\frac{y}{b}={{\tan }^{-1}}\left( \frac{x}{a}+{{\tan }^{-1}}\frac{y}{x} \right)\] \[\Rightarrow \] \[\tan \frac{y}{b}=\frac{x}{a}+{{\tan }^{-1}}\frac{y}{x}\] On differentiating both sides with respect to x, we get \[\frac{1}{b}{{\sec }^{2}}\left( \frac{y}{b} \right)\frac{dy}{dx}=\frac{1}{a}+\frac{1}{1+{{\left( \frac{y}{x} \right)}^{2}}}\times \frac{x\frac{dy}{dx}-y}{{{x}^{2}}}\] \[\Rightarrow \] \[\frac{1}{b}{{\sec }^{2}}\left( \frac{y}{b} \right)\frac{dy}{dx}=\frac{1}{a}+\frac{x\frac{dy}{dx}-y}{{{x}^{2}}+{{y}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\left\{ \frac{1}{b}{{\sec }^{2}}\left( \frac{y}{b} \right)-\frac{x}{{{x}^{2}}+{{y}^{2}}} \right\}=\frac{1}{a}-\frac{y}{{{x}^{2}}+{{y}^{2}}}\] \[\therefore \] \[\frac{dy}{dx}=\frac{\frac{1}{a}-\frac{y}{{{x}^{2}}+{{y}^{2}}}}{\frac{1}{b}{{\sec }^{2}}\left( \frac{y}{b} \right)-\frac{x}{{{x}^{2}}+{{y}^{2}}}}\]
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