Tangent plane to $f(x,y)=x^2ye^{1-xy}$

I wonder if my thoughts are correct when solving the following problem:

Find a tangent plane to the function $$f(x,y)=x^2ye^{1-xy}$$ at a point $((1,1),\;f(1,1))$. Express it both as a defining equation and parametric equation.

Clearly the given function can be represented using Taylor series expansion at the given point. The first two terms of this expansion give us linear approximation of the function at this point, which I guess will be exactly our desirable tangent plane.

So our plane using Taylor series $$z=f(1,1)+{\partial f\over \partial x}(1,1)(x-1)+{\partial f \over \partial y}(1,1)(y-1)\\ = 1+1\cdot (x-1) + 0\cdot (y-1)=x$$

I obtain the plane equation $z=x$. However I am not sure the result is correct. Is there a mistake in my logic?

Your answer is correct. The normal vector to the surface $$g(x,y,z)=x^2 y \exp (1-x y)-z=0$$ is its gradient, $$\hat{n}=\{\frac{\partial g}{\partial x},\frac{\partial g}{\partial y},\frac{\partial g}{\partial z}\}=\{2 x y e^{1-x y}-x^2 y^2 e^{1-x y},x^2 e^{1-x y}-x^3 y e^{1-x y},-1\}$$ at point $(1,1,1)$ (not that $f(1,1)=1$), i.e. $$\hat{n}=\{1,0,-1\}.$$ Therefore, the plane equations is, $$(1)x+(0)y+(-1)z=d \to x-z=d$$ Moreover, since point $(1,1,1)$ is on the plane, we have $$1-1=d\to d=0$$ Therefore, finally the plane equation is, $$x-z=0$$ In the parametric form the plane equation is, $$x(t)=t,\,z(t)=t$$ with arbitrary $y(t).$