[proofplan]
The inverse metric is characterized by the coordinate identity $g^{ik}g_{kj}=\delta^i_j$. We differentiate this identity with respect to the parameter $t$, using the product rule for the two matrix-valued component functions. Then we contract with the inverse metric to isolate $\partial_t g^{ij}$, and the Ricci flow formula follows by substituting $v_{ij}=-2\operatorname{Ric}_{ij}$.
[/proofplan]
[step:Differentiate the inverse metric identity in time]
Fix $t \in I$ and a point $p \in U$. In the chosen coordinate chart, the inverse metric components are defined by
\begin{align*}
g^{ik}(t,p)g_{kj}(t,p)=\delta^i_j,
\end{align*}
where $\delta^i_j$ denotes the Kronecker delta. Since the family $g(t)$ is smooth in $t$, both component functions $t \mapsto g_{ij}(t,p)$ and $t \mapsto g^{ij}(t,p)$ are differentiable. Differentiating the identity with respect to $t$ and applying the product rule gives
\begin{align*}
(\partial_t g^{ik})(t,p)g_{kj}(t,p)+g^{ik}(t,p)(\partial_t g_{kj})(t,p)=0.
\end{align*}
By the definition $v_{kj}=\partial_t g_{kj}$, this is
\begin{align*}
(\partial_t g^{ik})g_{kj}+g^{ik}v_{kj}=0.
\end{align*}
[guided]
Fix $t \in I$ and a point $p \in U$. The calculation is pointwise, so at this point we are simply differentiating the inverse of a positive definite matrix depending smoothly on $t$. In coordinates, the defining identity for the inverse metric is
\begin{align*}
g^{ik}(t,p)g_{kj}(t,p)=\delta^i_j,
\end{align*}
where $\delta^i_j$ is the Kronecker delta. The right-hand side is independent of $t$, so its $t$-derivative is zero.
Because $(g(t))_{t \in I}$ is a smooth one-parameter family of metrics, the component functions $t \mapsto g_{ij}(t,p)$ are smooth. The inverse matrix of a smoothly varying invertible matrix also has smooth entries, so $t \mapsto g^{ij}(t,p)$ is differentiable. Applying the product rule to the product $g^{ik}g_{kj}$ gives
\begin{align*}
\partial_t(g^{ik}g_{kj})
=
(\partial_t g^{ik})g_{kj}+g^{ik}(\partial_t g_{kj}).
\end{align*}
Since $\partial_t(\delta^i_j)=0$, we obtain
\begin{align*}
(\partial_t g^{ik})g_{kj}+g^{ik}(\partial_t g_{kj})=0.
\end{align*}
The tensor $v$ was defined by $v_{kj}:=\partial_t g_{kj}$, so the differentiated identity becomes
\begin{align*}
(\partial_t g^{ik})g_{kj}+g^{ik}v_{kj}=0.
\end{align*}
[/guided]
[/step]
[step:Contract with the inverse metric to isolate $\partial_t g^{ij}$]
Contract the identity
\begin{align*}
(\partial_t g^{ik})g_{kj}+g^{ik}v_{kj}=0
\end{align*}
with $g^{j\ell}$. Using $g_{kj}g^{j\ell}=\delta^\ell_k$, we get
\begin{align*}
(\partial_t g^{ik})\delta^\ell_k+g^{ik}g^{j\ell}v_{kj}=0.
\end{align*}
Thus
\begin{align*}
\partial_t g^{i\ell}=-g^{ik}g^{j\ell}v_{kj}.
\end{align*}
Now rename the free index $\ell$ as $j$ and the dummy index $j$ as $\ell$. Since the inverse metric is symmetric, $g^{\ell j}=g^{j\ell}$, this gives
\begin{align*}
\partial_t g^{ij}=-g^{ik}g^{j\ell}v_{k\ell}.
\end{align*}
This proves the variation formula for the inverse metric.
[/step]
[step:Substitute the Ricci flow variation]
If $g(t)$ evolves by Ricci flow, then by hypothesis
\begin{align*}
v_{k\ell}=\partial_t g_{k\ell}=-2\operatorname{Ric}_{k\ell}.
\end{align*}
Substituting this into the variation formula gives
\begin{align*}
\partial_t g^{ij}=-g^{ik}g^{j\ell}v_{k\ell}.
\end{align*}
Using $v_{k\ell}=-2\operatorname{Ric}_{k\ell}$, we obtain
\begin{align*}
\partial_t g^{ij}=-g^{ik}g^{j\ell}(-2\operatorname{Ric}_{k\ell}).
\end{align*}
Therefore
\begin{align*}
\partial_t g^{ij}=2g^{ik}g^{j\ell}\operatorname{Ric}_{k\ell}.
\end{align*}
By definition of the raised Ricci tensor components, $\operatorname{Ric}^{ij}:=g^{ik}g^{j\ell}\operatorname{Ric}_{k\ell}$, so
\begin{align*}
\partial_t g^{ij}=2\operatorname{Ric}^{ij}.
\end{align*}
This is the stated Ricci flow identity.
[/step]
