$$g(x,y,z) = x^2 + y^2 - z^2 - xyz = 12$$
Given a solution point $(x,y,z) = (3,4,1)$, write the first order Taylor Formula centered at $f(3,4)$ A unique continuous implicit function $z = f(x,y)$ exists.
The correct answer for the problem is $f(3,4) = 1$. However, I am not quite sure how to arrive at that. I have written down the working I have done, but I am not sure it's done the correct way. I do not need help with writing the Taylor formula, just finding $f(3,4)$.
$$g(3,4,1) = 12$$ $$f(3,4) = 3^2 + 4^2 - 0^2 - ((3)(4)(0)) - g(3,4,1) = 13-12 = 1$$
Please let me know if this working is correct. Thank you in advance
The first order approximation of $f(x,y)$ around $(3,4)$is given by $$ f(x,y) \approx f(3,4) + f'_x(3,4)(x-3) + f'_y(3,4)(y-4) $$
You already concluded that $f(3,4)=1$, now you must obtain the partial derivatives at $(3,4)$. You can either use the formula from the implicit function theorem, or derive the equality with respect to $x$ and $y$. For instance, deriving
$$ x^2+y^2-f(x,y)^2- x y f(x,y)=12 $$
with respect to $x$ gives you $$ 2x + 2 f'_x(x,y) f(x,y)-y(f(x,y) + x f'_x(x,y))=0. $$ so,
$$ f'_x(x,y)=\dfrac{y f(x,y) - 2x}{2f(x,y)-xy}. $$
Substituting $(x,y)=(3,4)$ yields $$ f'_x(3,4) = \frac{4\times 1 - 2\times 3}{2 \times 1 - 3\times 4 } = \frac 15 $$
Do the same for $f'_y(3,4)$ and you'll have your result.