Having trouble grasping the Implicit function theorem and its applications I'm trying to wrap my head around this theorem. I'd really appreciate if you can help me with this exercise and my pathetic attemp at solving it. (I'd also accept tips on how to approach these kinds of exercises in general.)

Consider the equation:
  $$(x+1)^2 - 3\cos y + \cos xz + \ln (x^2y^4 + 1) + e^{\sin z} - 3 = 0$$
(1) Prove that there's a neighborhood $U \subseteq \mathbb{R}^2$ of the point $(0,\pi/2)$ and an interval $I \subseteq \mathbb{R}$, $I=(\pi - \epsilon, \pi + \epsilon)$ such that the equation given defines implicitly a $C^1$ function $z=g(x,y): U \to I$.
(2) Find $\frac{\partial g}{\partial x}(0,\pi/2)$

I was taught to first consider the relevant level surface:
$$S=\{ (x,y,z) \in \mathbb{R}^3 : f(x,y,z) = (x+1)^2 - 3\cos y + \cos xz + \ln (x^2y^4 + 1) + e^{\sin z} - 3 = 0 \}$$
(My textbook speaks indistinctly of "equations" and "functions", but isn't the point of the theorem to determine when an equation defines a function? E.g., why would it be legitimate to assume a priori that $f(x,y,z)$ defines a function?)
Moving on, we know that
$$f(0,\pi/2,z) = e^{\sin z} - 1 = 0 \ \iff \ \sin z = 0$$
In particular, $(0,\pi/2, \pi) \in S$. We also know that $f$ is $C^1$ (because it's sum, composition, etc. of $C^1$ functions) and that $\frac{\partial f}{\partial z}(0,\pi/2, \pi) = -1 \neq 0$.
Therefore the Implicit function theorem guarantees that there's an open disk $B \subseteq \mathbb{R}$, an open neighborhood $V \subseteq \mathbb{R}^3$ of the point $(0,\pi/2,\pi)$ and a $C^1$ function $\phi: B\to \mathbb{R}$ such that $S \cap V = \text{Graph}(\phi)$. How can i get from this $\phi$ to the $g$ that I need to prove exists? They exist in different vector spaces, right?

Moving on to $(2)$...
$$\frac{\partial g}{\partial x}(0,\pi/2) = \frac{- \frac{\partial f}{\partial x}|_{(0,\pi/2, g(0,\pi/2))}}{\frac{\partial f}{\partial z}|_{(0,\pi/2, g(0,\pi/2))}} = 2$$
 A: I'm not sure if you made a typo or it's a genuine mistake, but the Implicit Function Theorem guarantees that there exists an open set $B \subseteq \Bbb{R}^2$ containing the point $(0,\pi/2)$, and an open set $V \subseteq \Bbb{R}^3$ containing $(0, \pi/2, \pi)$, and a $C^1$ function $\phi: B \to \Bbb{R}$, such that
\begin{equation}
S \cap V = \text{Graph}(\phi)
\end{equation}
So the function $\phi$ is actually the function $g$ that you seek.
By the way, I didn't verify your computations with the numbers, but the formula you used to compute the partial derivative is right.
Also, I'm not entirely sure what your initial concern regarding the use of terms "equation" and "function" is about. There are several equivalent ways of stating the IFT, but one way is to say that it is a criterion for checking when the level set of a $C^1$ function $f$ can locally be expressed as the graph of a $C^1$ function $g$. Also, we are not assuming a priori that your given function $f$ implicitly defines a function $g$ defined on a neighbourhood $U$ of $(0,\pi/2)$; this is something which has to be proven, and you have in fact proven it.
