Discussion
For \( z = f(x^{2} + y^{2}) \) with \( f \) arbitrary, eliminating \( f \) gives
Answer: B
With \( u = x^{2} + y^{2} \) the chain rule gives \( p = 2x f'(u) \) and \( q = 2y f'(u) \), so dividing removes the unknown derivative of \( f \):
\[ \frac{p}{q} \;=\; \frac{x}{y} \quad \Longrightarrow \quad yp - xq \;=\; 0. \]
The relation says the surface is unchanged along circles about the axis, which is exactly what dependence on \( x^{2} + y^{2} \) means
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