Finding a tangent plane to the surface $g(x,y,z)=0$

The tangent plane writes $$ g_x(x_0,y_0,z_0) (x-x_0)+ g_y(x_0,y_0,z_0) (y-y_0)+ g_z(x_0,y_0,z_0) (z-z_0) $$

It is convenient to introduce the variables $u=xyz,v=x^2+y^2+z^2$ so that $g(x,y,z)=f(u,v)$ and $g(x_0,y_0,z_0)=f(u_0,v_0)$.

It follows $$ du = yz dx + xz dy + xy dz, dv = 2x dx + 2y dy + 2z dz, $$

From \begin{eqnarray*} df &=&\frac{\partial f}{\partial u}du+ \frac{\partial f}{\partial v}dv \\ &=& \left( \frac{\partial f}{\partial u} yz+ \frac{\partial f}{\partial v} 2x \right) dx + \left( \frac{\partial f}{\partial u} xz+ \frac{\partial f}{\partial v} 2y \right) dy + \left( \frac{\partial f}{\partial u} xy+ \frac{\partial f}{\partial v} 2z \right) dz \end{eqnarray*} we deduce \begin{eqnarray*} g_x(x_0,y_0,z_0)&=& f_u(u_0,v_0) y_0z_0+ f_v(u_0,v_0) 2x_0 \\ g_y(x_0,y_0,z_0)&=& f_u(u_0,v_0) x_0z_0+ f_v(u_0,v_0) 2y_0 \\ g_z(x_0,y_0,z_0)&=& f_u(u_0,v_0) x_0y_0+ f_v(u_0,v_0) 2z_0 \end{eqnarray*} From here, you can easily conclude.