Plane equation to $f(x,y)=\frac{1}{1-x^2+y}$, to the point $(x_0,y_0)=(1,2)$

Question

My solution:

$f'_x=\frac{2xy}{(1-x^2y)^2}$ and $f'_y=\frac{x^2}{(1-x^2y)^2}$

by substituting $x_0,y_0=(1,2)$ I get that $f'x=\frac{2 \cdot 1 \cdot 2}{(1-1^2 \cdot 2)^2}=4$ and $f'y=\frac{1^2}{(1-1^2 \cdot 2)^2}=1$ and $z_0=f(1,2)=\frac{1}{1-1^2+2}=\frac{1}{2}$

for the equation of the plane:

$f'_x(x_0,y_0)(x-x_0)+f'_y(x_0,y_0)(y-y_0)-(z-z_0)=0$

by substituting the values above:

$4(x-1)+1(y-1)-(z-\frac{1}{2})=0$ which is equal to:

$4x+y-z=4.5$

Is my solution correct?

Answer 1

No, because$$\frac{\partial f}{\partial x}=\frac{2x}{(1-x^2+1)^2}\text{ and }\frac{\partial f}{\partial y}=\frac{-1}{(1-x^2+y)^2}.$$