(h@f@g)(x) is known as a composite function. Here's how composite functions work:
let's say that x=1. Your function g(x) = 1/3(x) now produces a y output of 1/3 , since g(1)=1/3(1). Within a composite function, the y-value of one function becomes the x-value of the next, like so:
g(1)=1/3=>f(1/3)=17/3=>h(17/3)=17
Therefore,
(h@f@g)(1)=17
Based on this, to find the function for (h@f@g)(x) (combined from right to left, by the way), simply replace x in f(x) with the function g(x), and replace x in h(x) with the function of (f@g)(x), to get (h@f@g)(x).
This, simplified, is equal to 2x+15, and therefore (h@f@g)(1)=2(1)+15=17
Hope that helps!