So, you are given that g is continuous. This means that for any epsilon>0, there exists a delta>0, such that |a-x|<delta implies |g(a)-g(x)|<epsilon. You want to prove a similar implication for f, that is, for a given epsilon>0, you need the existence of a delta>0, such that |a-x|<delta implies |f(a,b)-f(x,b)|<epsilon (where b is arbitrary).

However, by definition of f, we have that |f(a,b)-f(x,b)| = |g(a)-g(x)|. Let me know, if you can see where this goes.