Knowing the equation of the tangent plane $\pi: 2x + y + 3z = 0$ at the point $(1, 1, 1)$ of a function $f(x, y)$, how can I find the values of $f'_x(1, 1)$ and $f'_y(1, 1)$?
I thought of this: since I know the equation of the plane, then I know it comes from
$$\pi: z=f(1, 1) + f'_x(1, 1)(x-1) + f'_y(1, 1)(y-1)$$
Thence by comparison I shall say, since $z = -\frac{2}{3}x - \frac{1}{3}y$
$$0 = f(1, 1) - f'_x(1, 1) - f'_y(1, 1)$$
and most of all
$$f'_x(1, 1) = -\frac{2}{3} \qquad f'_y(1, 1) = -\frac{1}{3}$$
Is this correct?
The solution is correct but I'd add perhaps a little bit more rigor.
Since $(1,1,f(1,1))$ touches the plane we have $f(1,1)=-1\,.$ The equation you labelled with $\pi$ holds for all $x,y,z$ in the plane. Choosing two points $(2,0,0)$ and $(2,2,-2)$ in the plane gives us a system \begin{align} -\tfrac 43&=-1+f'_x-f'_y\,,\\[2mm] -2&=-1+f'_x+f'_y \end{align} that has the unique solution $f'_x=-2/3\,,f'_y=-1/3\,.$