Levi-Civita connection for a metric in $\mathbb{R}^{3}$ 
Let $g$ be a metric in $\mathbb{R}^{3}$ defined as $\partial_{x}, \partial_{y}, \partial_{z}$ are orthogonal everywhere, and $g(\partial_{y},\partial_{y})=1, g(\partial_{z},\partial_{z})=f(x), g(\partial_{x},\partial_{x})=f(x)$ for $f$ being a positive function. Compute $\nabla_{\partial_{i}}\partial_{j}$ for all $i,j$ being $x,y$ or $z$ and $R^{\nabla}$ (the Riemannian curvature tensor). Moreover, show that for $M=\{x=c\}$, the restricted metric is isometric to $\mathbb{R}^{2}$.

For the first question, this is just computing the Levi-Civita connection for the metric $g$. This can be computed with the Cristoffel symbols for the metric. However, I'm not sure on how to compute them as we are not given a metric of the form $u dx^{2}+v dy^{2}+t dz^{2}$. What would be a way of writing the metric in this form? Or would it just be easier to compute it using an index-free argument? I'm not sure on how to proceed since we haven't covered much of this in class.
For the second question, since $R^{\nabla}$ is tensorial in all slots, it suffices to compute $R^{\nabla}(\partial_{i},\partial_{j})\partial_{k}$ for all $i,j,k$ being $x,y,z$. Is this correct? If so, would it just be $\nabla_{\partial_{i}}\nabla_{\partial_{j}}\partial_{k}-\nabla_{\partial_{j}}\nabla_{\partial_{i}}\partial_{k}$ since the Lie bracket $[\partial_{i},\partial_{j}]$ vanishes? For the final question, I think this would be equivalent to showing that $g(\partial_{i},\partial_{j})=\langle \partial_{i}, \partial_{j} \rangle_{Euc}$. How would I go about this? Thank you for the help.
 A: Once you work in the global coordinates $(x,y,z)$ this is just a computation together with a thorough understanding of the definitions.
First, the usual proof of the existence of the Levi-Civita connection gives the coordinate formula for the Christoffel symbols
\begin{align} \label{a}\tag{1}
\Gamma_{ij}^k = \frac12 g^{kl}(\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij})
\end{align}
where I have adopted the usual summation convention, identified $g$ with its matrix in these coordinates and written $g^{kl} = (g^{-1})_{kl}$. 
The information you're given about $g$ says precisely that in the obvious coordinates on $\mathbb{R}^3$, $g$ has matrix representation
$$g_{(x,y,z)} = \begin{bmatrix}
f(x)&0&0& \\0&1&0& \\0&0&f(x)&
\end{bmatrix}.$$
From here, finding $\Gamma_{ij}^k$ is just a computation using (\ref{a}). You should get the non-zero Christoffel symbols to be
\begin{align}
\Gamma_{ij}^k(x,y,z) = \begin{cases} -\frac{\partial_xf(x)}{f(x)}  \quad \text{ if } i = j \in \{1,3\} \text{ and } k = 1, \\ \\
\frac{\partial_xf(x)}{f(x)} \quad \text{ if } \{i,j\} = \{1,3\} \text{ and } k = 3.
\end{cases}
\end{align} 
Computing the curvature tensor is again just a computation in the global coordinates using the identity
$$R = R_{ijk}^{\,\,\,\,\,l} dx^i \otimes dx^j \otimes dx^k \otimes \partial_l$$
where $R_{ijk}^{\,\,\,\,\,l}$ satisfies $R_{ijk}^{\,\,\,\,\,l} \partial_l = R(\partial_i,\partial_j) \partial_k$. As you noticed, this computation is made easier by the fact that the Lie brackets $[\partial_i, \partial_j]$ all vanish and hence is straightforward (if lengthy) now you know the Christoffel symbols.
We are left to see that $M = \{x=c\}$ is isometric to $\mathbb{R}^2$. First, let's see what the induced metric on $M$ is. It is clear that $(c,y,z) \mapsto (y,z)$ is a global coordinate chart on $M$ and that $\operatorname{Id}:M \to \mathbb{R}^3$ is an immersion. Let $\tilde{g} = \text{Id}^* g$ be the induced metric on $M$. It is not too hard to check that $\tilde{g}$ has matrix
$$\tilde{g}_{(c,y,z)} = \begin{bmatrix} 1&0\\0&f(c) \end{bmatrix}$$ with respect to our global coordinates on $M$. 
Notice here that $g$ and $\langle \cdot, \cdot \rangle_{\text{Euc}}$ act on different spaces so one can only write $g(\partial_i, \partial_j) = \langle \partial_i, \partial_j \rangle_{\text{Euc}}$ up to some identification between the spaces - i.e. if we are thinking of $\partial_k$ on the RHS as really meaning $d\Phi (\partial_k)$ for some diffeomorphism $\Phi: M \to \mathbb{R}^2$. I assume that your idea was to take $\Phi(c,y,z) = (y,z)$ but it is clear from the matrix form of $\tilde{g}$ that this won't work.
Instead, it is natural to define the map $\Phi: M \to \mathbb{R}^2$ by $\Phi(c,y,z) = (y, \sqrt{f(c)} z)$. It is immediate that $\Phi$ is a diffeomorphism so we want to check that $\tilde{g} = \Phi^* \langle \cdot, \cdot \rangle_{\text{Euc}}$. For this, it is enough to check that $\Phi^* \langle \cdot, \cdot \rangle_{\text{Euc}}$ has the right matrix representation. 
This is another computation. We have, for example,
\begin{align}
\Phi^* \langle \cdot, \cdot \rangle_{\text{Euc}}(\partial_z |_p, \partial_z |_p) = \langle d\Phi_p (\partial_z |_p), d\Phi_p (\partial_z |_p) \rangle_{\text{Euc}}
\end{align}
Now, it isn't too hard to see that for $F \in C^\infty(\mathbb{R}^2)$ and $p = (c,y,z)$ we have
$$d\Phi_p (\partial_z |_p)(F) = \partial_z |_p (F \circ \Phi) =  \sqrt{f(c)} \partial_z|_{\Phi(p)} F.$$
Therefore $d\Phi_p (\partial_z|_p) = \sqrt{f(c)}\partial_z|_{\Phi_p}$ and so 
$$\Phi^* \langle \cdot, \cdot \rangle_{\text{Euc}}(\partial_z |_p, \partial_z |_p) = f(c) = \tilde{g}(\partial_z |_p, \partial_z |_p).$$
The other entries in the matrix follow similarly. 
