Let $e=uv$ be an edge of $G$.
Let $f$ and $g$ be the other two edges incident with $v$.
Let $G'=G-f-g$.
Claim 1: $G'-v$ is 2-connected.
$G'-v$ is the same as $G-v$.
Claim 2: $G'$ has a perfect matching.
You can prove this by showing directly that the Tutte condition holds.
I assume that you are familiar with the standard proof of Petersen's theorem,
so I will not elaborate on every detail.
Note that $G'$ has only 2 vertices of even degree.
Let $S$ be a vertex subset of $G$, $o(G'-S)$ the number of odd components of $G'-S$.
At most one odd component can contain $v$; that component can have as few as 1 edge to $S$ (since $G'$ is still connected).
All other odd components contain a vertex different from $v$, so because $G'-v$ is 2-connected all other odd components have at least 2 edges to $S$.
At most two odd components can contain a vertex of even degree (even in $G'$).
All other odd components must have at least three edges to $S$ (as in the
standard proof of the Petersen theorem).
Let $x$ be the number of edges between $S$ and the odd components of $G'-S$.
We have just shown that $x\geq 3o(G'-S)-4$.
On the other hand $S$ can accomodate at most $3|S|$ edges, so $x\leq3|S|$.
This means that $o(G'-S)\leq|S|+\frac{4}{3}$.
Now finally use the fact that $G$ is even, so $o(G'-S)$ and $|S|$ have the same parity.
This shows that $o(G'-S)\leq|S|$ and the Tutte condition holds.
Final argument: the perfect matching of $G'$ is a perfect matching of $G$ and it contains $e$,
since the degree of $v$ in $G'$ is 1.
Well, you would be done if you could extend every set $S$ of $k-1$ independent edges to a set of $k$ independent edges [make sure you see why]. And in fact you can.
Let $M$ be a perfect matching. Then $M \cup S$ contains at least one path or cycle $L$ with an odd number $\ell$ of edges such that only $\lfloor \frac{\ell}{2}\rfloor$ of those edges are in $S$ and the remaining $\lfloor \frac{\ell}{2}\floor +1$ are in $M$. So let $S' = S - (S\cap L) + (M \cap L)$. Then (a) $S'$ is an independent set with $k-1+1 =k$ edges, and (b) every vertex covered by $S$ is also covered by $S'$ [make sure you see why for both (a) and (b)].
Then, as $G$ has a perfect matching and at least $2k+2$ vertices, a perfect matching in $G$ has $k+1$ edges. Because $G$ also has the property that every independent set of $k$ edges can be extended to a perfect matching, this implies that there is a perfect matching containing $S'$ that has at least one edge $e$ not in $S'$. Then $\{e\}+S'$ is an independent set of edges. Therefore, as every vertex covered by $S$ is also covered by $S'$, it follows that $\{e\}+S$ is also an independent set of edges with $e$ not be in $S$ [make sure you see why]. So $S$ indeed can be extended to an independent set of $k$ edges, giving the desired result.
Best Answer
To apply Tutte-Berge, we want to find a small set $U \subseteq V$ such that $G[V-U]$ has many odd components. There are two limitations on the number of odd components:
Based on the second constraint, we can consider two kinds of odd components. Odd components of $G[V-U]$ with $1$ or $3$ vertices must have at least $3$ edges to $U$ (just by counting the maximum number of edges within a component), and odd components with at least $5$ vertices still must have at least $1$ edge to $U$ (since you can't have an odd component by itself in a $3$-regular graph).
Let $u = |U|$, let $x$ be the number of small odd components, and let $y$ be the number of large odd components. Then the two constraints are $$x + 5y \le n-u \qquad \text{and} \qquad 3x + y \le 3u.$$
Combining these with weights $\frac18$ and $\frac38$, we get $\frac54x + y \le \frac18n + u$, so $x + y \le \frac18 n + u$ as well. By the Tutte-Berge formula, the number of vertices covered by a maximum matching is the minimum of $n+u-x-y$ over all sets $U$, and we've shown that this always satisfies $$ n + u - x - y \ge n + u - (\tfrac18 n + u) = \tfrac78 n $$ So the matching covers at least $\frac78n$ vertices.