A question on the implicit function theorem.

Question

The question is given below:

Approximate by a second-degree polynomial the solution of $z^3 + 3xyz^2 - 5x^2y^2z + 14 = 0$, for $z$ as a function of $x,y,$ near (1, -1, 2).

Could anyone give me a hint for the solution please?

My ideas:

our desired function will map (x,y) to z, so $n=2$ and $m = 1$. further $x_{0} = (1, -1)$ and $y_{0} = 2$, and $F(x,y,z) = z^3 + 3xyz^2 - 5x^2y^2z + 14.$ clearly $F$ has a continuous partial derivative and the derivative is $$DF(x,y,z)[3yz^2 -10xy^2z \quad 3xz^2-10x^2yz \quad 3z^2+6xyz-5x^2y^2 ],$$

And hence,

$$DF(1,-1,2)=[-32 \quad 32 \quad -5 ],$$

Then,

$$DF(1,-1,2)\begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix}= -5z,$$

But then what next?

Answer 1

You don't need the derivative of $F$. The steps I suggest you take are the following:

1. We know, from the implicit function theorem, that, near the point $(1, -1)$, there exists a function $z:\mathbb R^2\to \mathbb R$ such that $$z(x,y)^3 + 3xyz(x,y)^2 - 5x^2y^2z(x,y) + 14 = 0.$$ That is, in this step, we write $z$ as the function. We also know, of course, that $z(1,-1)=2$.
2. Now, take the equation above, and take the derivative, with respect to $x$, of it. Be careful: $z$ is a function of $x$ now, but $y$ is not! So, $\frac{\partial z(x,y)^3}{\partial x} = 3z(x,y)^2\cdot \frac{\partial z(x,y)}{\partial x}$, for example.
3. From the point above, you should get an equation from which you can extract the value of $\frac{\partial z}{\partial x}$ at $(x,y)=(1,-1)$.
4. Do the same to get the partial derivative for $y$.