Remember the chain rule.
L(x)=f(g(x))
L'(x)=f'(g(x))g'(x)
take the derivative of f(g(x)). just treat them like they are variables. so you get:
h'=f'(g(x))g'(x)
now plug in your x value and evaluate:
h'(1)=f'(g(1))(g'(1))
substitute in values that you know and evaluate again
h'(1)=f'(3)(-3)
h'(1)=(-5)(-3)=15
