Is there a method of knowing which of the Christoffel symbols of second kind survive and vanish for a given metric? I'm trying to solve a problem and the given $g_{ij}$ metric is $$\left[\begin{array}{cc}1&0&0\\0&(x^1)^2&0\\0&0&(x^1 \sin x^2)^2\end{array}\right]$$
The non-zero Christoffel symbols of the second kind, $\Gamma^{i}_{\;\,jk}$, for this problem happen to be:
$i=1$: $\Gamma^{1}_{\;\,22}$, $\Gamma^{1}_{\;\,33},$
$i=2$: $\Gamma^{2}_{\;\,12} =\Gamma^{2}_{\;\,21}$, $\Gamma^{2}_{\;\,33},$
$i=3$: $\Gamma^{3}_{\;\,13} =\Gamma^{3}_{\;\,31}$, $\Gamma^{3}_{\;\,23} =\Gamma^{3}_{\;\,32}$.
My question is, how can one be sure these are the only non-zero ones? It's difficult to check all the possible combinations of $ijk$, especially if there are time constraints.
 A: Christoffel symbols are given by $\Gamma_{ij}^{k}=\frac 1 2  g^{km}(g_{mj,i}+g_{mi,j}-g_{ij,m})$.
I will give a method for diagonal metrics, as the one you have here. I think that for non-diagonal metrics there is no way to simplify significantly the calculation.
We may first notice that for a diagonal metric, only terms with $k=m$ survive. So we may look at Christoffel symbols as  $\Gamma_{ij}^{k}=\frac 1 2  g^{kk}(g_{kj,i}+g_{ki,j}-g_{ij,k})$ which simplifies the work since now there is no summation on $m$ - there is only this one term. 
Now, because the metric is diagonal, only the Christoffel symbols with two equal indices survive. That is, with $i=j$, $i=k$ or/and $j=k$. Otherwise, all terms in the parentheses vanish. And even if two indices are equal, say $i$ appears twice in the symbol, we need to check whether $g_{ii}$ is dependent on the third index appearing in the symbol. If not, the symbol vanishes. In particular, if $g_{ii}$ is constant, all Christoffel symbols where $i$ appears twice (or more), vanish.
Sidenote (not really important): we notice that if the metric in independent of some variable, a term containing a derivative with respect to it always vanishes, so these indices appear in the symbol only twice or not at all (check that you understand why). In your example, $3$ indeed appears only in pairs, as the metric is independent of $x^3$.
To summarize: in order not to vanish, the Christoffel symbol needs to have two identical indices, where the matching metric element is dependent on the third. (For example, a Christoffel symbol $\Gamma_{ij}^j$ doesn't vanish if and only if $g_{jj}$ is dependent on $x^{i}$). A method of working this out in an efficient way is to write down un-ordered triples of indices fulfilling this, and then let them appear in the Christoffel symbol in all three possible ways.
This way we can quickly decide which Christoffel symbols vanish and which need a calculation.
For example, let's take your case:
After finding which elements in the metric depend on what, we are left with the triples of indices $(2,2,1)$, $(3,3,1)$ and $(3,3,2)$, (since $g_{11}$ is constant, $g_{22}$ depends only $x^1$ and $g_{33}$ depends on $x^1$ and $x^2$) appearing in any possible order in the symbol.
And indeed, as you got, the only Christoffel symbols that don't vanish in your case are those with these triples of indices: $\Gamma_{22}^{1}$, $\Gamma_{21}^{2}=\Gamma_{12}^{2}$, $\Gamma_{33}^{1}$, $\Gamma_{31}^{3}=\Gamma_{13}^{3}$, $\Gamma_{33}^{2}$ and $\Gamma_{32}^{3}=\Gamma_{23}^{3}$.
A: CLARIFICATION: In general, computing Christoffel symbols is quite painful and usually can be avoided. Not only do you have to compute every partial derivative of every component of the metric tensor, you also have to compute the inverse tensor. The following is for a diagonal metric tensor only.
Suppose $g_{ij} = a_i\delta_{ij}$.
\begin{align*}
\partial_kg_{ij} &= \partial_ka_i\delta_{ij}\\
2g_{kp}\Gamma^p_{ij} &= \partial_ig_{kj} + \partial_jg_{ik} - \partial_kg_{ij}\\
&= \partial_ia_k\delta_{jk} + \partial_ja_k\delta_{ik} - \partial_ka_i\delta_{ij}\\
2\Gamma^k_{ij} &= a_k^{-1}(\partial_ia_k\delta_{jk} + \partial_ja_k\delta_{ik} - \partial_ka_i\delta_{ij}).
\end{align*}
This is zero if $i, j, k$ are different from each other. Since the Christoffel symbol is symmetric in $i$ and $j$, it now suffices to consider the following cases:
If $i = j = k$, then
$$
2\Gamma^i_{ii} = a_i^{-1}\partial_ia_i.
$$
If $i= j \ne k$, then
$$
2\Gamma^k_{ii} = -a_k^{-1}\partial_ka_i.
$$
If $i \ne j = k$, then
$$
2\Gamma^k_{ik} = a_k^{-1}\partial_ia_k.
$$
