Find plane equation based on tangent plane equation (both planes are parallel)? I' m asked to find the plane equation for (x,y,z)-> 0,0,0.
I' m given the following function: $$f(x,y)=e^{x^{2}}+\ln(\frac{1}{xy})$$
Also the task specifies that the origin plane is parallel with the tangent plane and passes through the following point (x,y,z)->(1,1,e) (tangent plane)
I've found the tangent plane equation (after doing the partial derivatives) and the equation is :$$x(2e-1)+y-z-(e+2)=0$$ (Implicit form).
How do I find the plane passing through (0,0,0), origin,  given that I have now the tangent plane equation? Can I just copy the coefficients and write 0 at the end since they are parallel?
Like this perhaps: $$x(2e-1)+y-z=0$$

Your approch is right, and but I got different numbers: the plane tangent at $P=(1,1,f(1,1))=(1,1,e)$ should be
$$F_x(P)(x-1)+F_y(P)(y-1)+F_z(P)(z-e)=0\quad \Rightarrow\quad (2e-1)x-y-z-(e-2)=0$$
where $F(x,y,z)=z-e^{x^2}+\ln(xy)$ and
$(F_x,F_y,F_z)=(-2xe^{x^2}+1/x,1/y,1)$.
Hence the parallel plane through $(0,0,0)$ is $(2e-1)x-y-z=0$.
