Rewriting the equation:
\[
\frac{dy}{dx} = \frac{1 + y^2}{\tan^{-1} y - x}.
\]
Use substitution \( z = \tan^{-1} y - x \), then differentiate and solve. The solution is:
\[
z = C, \quad \text{where } z = \tan^{-1} y - x.
\]
Thus:
\[
\tan^{-1} y - x = C.
\]