This is easy and comes from the fact that $V$, being a Fréchet space, is barrelled, that is, the locally convex topology of $V$ coincides with its strong topology $\beta(V,V')$ (where $V'$ = topological dual ov $V$), which is the one induced by the barrels = closed, absolutely convex and absorbing subsets of $V$, since by the bipolar theorem these are precisely the polars of pointwise bounded subsets in $V'$. In other words, in a barrelled Hausdorff locally convex vector space (henceforth lcs for short) every barrel is a neighborhood of zero. Equivalently, by e.g. Proposition 11.1.1, pp. 219 of the book by H. Jarchow, Locally Convex Spaces (B. G. Teubner, 1981), a Hausdorff lcs $V$ is barrelled if and only if for every Hausdorff lcs $W$ any pointwise bounded subset of $\mathscr{L}(V,W)$ ( = space of continuous linear maps from $V$ into $W$) is equicontinuous. This is the key result we will use.
Let now $T:X\times V\rightarrow W$ satisfy your hypotheses. For any compact subset $K\subset X$, we have that $T(K,v)\subset W$ is compact, hence bounded, for all $v\in V$. In other words, $T(K,\cdot)\subset\mathscr{L}(V,W)$ is pointwise bounded, hence equicontinuous. Finally, let $(p,v)\in X\times V$ and $U\subset W$ open with $T(p,v)\in U$. Since $T(\cdot,v)$ is continuous, there is a compact neighborhood $K\subset X$ of $p$ (since $X$ is locally compact) such that $T(K,v)\subset U$. Since $T(K,\cdot)$ is equicontinuous, there is $U'\subset V$ open with $v\in U'$ such that $T(K,U')\subset U$. We are done. Notice that it is not necessary to assume that $X$ is metrizable, and $W$ can be any Hausdorff lcs.
We remark that a similar argument, with the help of the Banach-Steinhaus theorem, can be used to prove joint continuity of separately continuous bilinear maps from metrizable lcs $E,F$ into a Hausdorff lcs $G$ if $E$ or $F$ is barrelled.