Normal to a surface of revolution meets its axis

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Prove that the normal of the surface of revolution

z=f(x2+y2)(f′≠0) z=f\left(\sqrt{x^{2}+y^{2}}\right) \quad (f' \neq 0)

meets the axis of rotation.

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Find the point of intersection, with the axis of rotation, of the normal at an arbitrary point (x0;y0;z0)\left(x_{0} ; y_{0} ; z_{0}\right) of the surface of revolution

z=f(x2+y2) z = f\left(\sqrt{x^{2}+y^{2}}\right)

where f(u)f(u) is a differentiable function, f′(u)≠0f^{\prime }(u) \neq 0.

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References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Функции многих переменных. 2-е изд., ФИЗМАТЛИТ, 2003.