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For easy typing, we let f = \phi in solution. Note that since fhas a strictly increasing derivative on (a,\xi)\cup(\xi,b)it is enough to check the convexity inequality on \xi.
Since \xi \in (a,b), for t \in [0,1], we can write \xi = ta + (1-t)b.
Since f' exists and finite, by Mean Value Theorem, there exists m,n \in (a,b)
 such that f'(m) = \frac{f(\xi) - f(a)}{\xi - a} ~\text{and} ~ f'(n) = \frac{f(b) - f(\xi)}{b - \xi}.(here we assume that m<n)
Since f' increasing, we have f'(m) \le f'(n). Then
f(\xi) = f(a) + f'(m)(\xi - a) \le f(a) + f'(n)(\xi - a) = f(a) + \frac{f(b) - f(\xi)}{b - \xi}(\xi - a).
Note that b-\xi = t(b-a) ~\text{and} ~ \xi - a = -(b-a)(t-1), the computation part is easy and left it to reader.
Multiply b-\xi = t(b-a)to both sides, then we getf(\xi)t(b-a) \le f(a)t(b-a) + [f(b) - f(\xi)](-(b-a)(t-1)).
After simplify, we get f(\xi)(b-a) \le t(f(a) - f(b)(b-a) - (t-1)f(b)(b-a) \implies f(\xi) \le tf(a) + (1-2t)f(b).
Since t \in [0,1], 1-2t \le 1-t. So, we have f(\xi) = f(ta + (1-t)b) \le tf(a) + (1-t)f(b).
So, f = \phi is convex on (a,b).