$f(x,y,z)=x^3-2y^2+z^2$, $x_0=(1,1,1)$ and $g(1,1)=1$, $f(x,y,g(x,y))=0$
find $g_x(1,1), g_y(1,1)$ using chain rule, are there any other ways to find it?
my attempt at it with respect to x, I think is not very good,
$$x^3-2y^2+(g(x,y))^2=0 \rightarrow 3x^2+2(g(x,y))g_x(x,y)=0$$
Then solve for $g_x(1,1)$? What about the implicit function theorem? Many thanks
Given: $${x_0} = (1,1,1),f(x,y,z) = {x^3} - 2{y^2} + z$$ with $$f({x_0}) = 0$$ and $$\nabla f = (3{x^2}, - 4y,1)$$ with $$\nabla f({x_0}) = (3, - 4,1) \ne \vec 0$$ Then, implicit fuction theorem says: Near ${x_0}$
$$\begin{gathered} f(x,y,g(x,y)) \equiv 0 \hfill \\ {x^3} - 2{y^2} + g(x,y) \equiv 0 \hfill \\ g(x,y) = - {x^3} + 2{y^2} \hfill \\ \end{gathered}$$ so: $$\begin{gathered} \nabla g = ( - 3x,4y) \hfill \\ \frac{{\partial g}}{{\partial x}}(x,y) = - 3x,\frac{{\partial g}}{{\partial y}}(x,y) = 4y \hfill \\ \end{gathered} $$ with $$\begin{gathered} \frac{{\partial g}}{{\partial x}}(1,1) = - 3 \hfill \\ \frac{{\partial g}}{{\partial y}}(1,1) = 4 \hfill \\ \end{gathered} $$