 Solution

Write the surface as the level set $F(x,y,z)=x^2{y}^3-y+1-z=0$. The gradient is normal to a regular level surface because differentiating $F$ along any surface curve gives $\nabla F\cdot \mathbf{v}=0$ for its tangent vector. At the specified point,Thus a normal vector and a unit normal are $(2,2,-1)$ and $(2,2,-1)/3$, respectively. The tangent-plane displacement is perpendicular to that normal, givingReversing the normal's sign gives the same plane.