= Solution
Write the surface as the level set $F(x,y,z)=x^2y^3-y+1-z=0$. 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,
$$
\nabla F=(2xy^3,3x^2y^2-1,-1)=(2,2,-1).
$$
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, giving
$$
\boxed{\mathbf n=(2,2,-1),\qquad 2(x-1)+2(y-1)-(z-1)=0,\quad\text{or}\quad z=2x+2y-3.}
$$
Reversing the normal's sign gives the same plane.
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