#f(x)=1-x#
#g(x)=x^2#
#h(x)=1/x#
To find #f(h(g(x)))# we start from the innermost bracket, and then apply each function to what we have 'so far' as though it were the #x# value in that function.
So, starting from the innermost parentheses, we have #g(x)=x^2#. We don't do anything else with that at this stage, but now we are going to find #h(g(x))#.
#h(x)=1/x#, but in this case we let #g(x)#, which is #x^2#, stand in for the #x# in this expression, so we end up with #h(g(x))=1/x^2#.
The next step is to find #f(h(g(x)))#, our answer. #f(x)=1-x# but we let #h(g(x))=1/x^2# stand in for #x#, so we end up with #f(h(g(x)))= 1 - 1/x^2#
