Let $V$ be a complex vector space, which we endow with the finest linear topology. Then continuous dual $V'$ coincides with the algebraic dual $V^*$. We choose the weak-star topology $\sigma(V^*,V)$ on $V^*$. Consider the map $(-,-) \ : \ V^* \times V \to \mathbb C$ defined by $(\phi,v)=\phi(v)$. I have been able to show that it is separately continuous and jointly sequentially continuous. I would like to know if it is jointly continuous.
No, whenever $V$ is infinite dimensional the evaluation isn't continuous with respect to the product of the weak$^*$ and the finest locally convex topology. Indeed, continuity at $0$ would give a finite set $E\subseteq V$, $\varepsilon >0$, and a $0$-neighbourhood $U$ in the finest locally convex topology such that $|f(u)|\le 1$ whenever $|f(e)|\le \varepsilon$ for all $e\in E$ and $u\in U$. Given $v\in V$ not contained in the linear span of $E$ you can choose $\delta>0$ such that $\delta v \in U$ and a linear functional $f$ which vanishes on $E$ with $|f(v)|\ge 2/\delta$ which yields a contradiction.
The argument in my answer also works for the finest linear topology. You just need that neighbourhoods of zero are absorbing.