Multivariable calculus/ How to solve problems involving implicit definition of a function in a function?

Question

Hello all math geniuses out there here is a question that probably does require much thinking, but i can't seem to figure out.

The equation $x^3 +yz-xz^2=0$ implicitly defines a function $z=f(x,y)$ for which $f(-2,1)=2$. Calculate $f_x(-2,1)$

Here is how i think about it $$f_x= 3x^2-z^2$$ $$f_y= z$$ $$f_Z=y-2xz$$ should I just evaluate everything at $(-2,1)$ or what?

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Honestly every answer below is satisfying and i dont know who to give best answer for so i am just gonna give it to the first one. The rest thank you for your time

Answer 1

$$ x^3+yf(x,y)-xf(x,y)^2=0 \text{\\Differentiate with respect to x}\\ 3x^2+yf_x(x,y)-f(x,y)^2-2xf(x,y)f_x(x,y)=0 $$ Plug in (-2,1): $$ 12+1\cdot{f_x(-2,1)}-4+8f_x(-2,1)=0\\ 9f_x(-2,1)=-8\\ f_x=-\frac{8}{9} $$

Answer 2

The equation is quadratic in $z$, there are two possible values of $z$ when plugging in for the derivatives.

$$ 2 z = y/x \pm \sqrt{(y/x)^2+ 4 x^2} $$

and evaluate everything at:

$$ (-2,1,z_1), (-2,1,z_2) $$

Answer 3

Rewrite the equation as

$x^3 + y*f(x,y) -x*f^2(x,y) = 0$

Take derivative wrt x:

$3x^2 + y*f_x(x,y) - (1*f^2(x,y) + x*2*f(x,y)*f_x(x,y)) = 0$

Now you can calculate $f_x(-2,1)$