You’re missing the basic idea behind composition. Let’s look just at the first step, $(g\circ h)(x)=g\big(h(x)\big)\;,$ where $g(x) = -5x^3$ and $h(x) = -4x^2+10$. The notation $g\big(h(x)\big)$ means that you’re going to use the output of $h$ as the input to $g$. You feed $x$ into $h$, and you $-4x^2+10$ as output. Now you feed that output into $g$:
$g\big(h(x)\big)=g\left(-4x^2+10\right)\;.$
What does $g$ do to its input? When the input is simply $x$, it spits out $-5x^3$. In other words, it cubes the input and then multiplies it by $-5$. When the input is $-4x^2+1)$, it’s going to do the same thing: cube it, and then multiply by $-5$. Thus,
$g\big(h(x)\big)=g\left(-4x^2+10\right)=-5\left(-4x^2+10\right)^3\;.\tag{1}$
When you’re just starting to work with composition of functions it may help to introduce new variables temporarily. Here, for instance, I could let $u=-4x^2+10$. Then
$g\left(-4x^2+10\right)=g(u)\;,$
and it may be easier to see that $g(u)=-5u^3=-5\left(-4x^2+10\right)^3$ than to see it all in one go as in $(1)$.
Now we want to feed $g\big(h(x)\big)$ into $f$:
$f\left(g\big(h(x)\big)\right)=f\left(-5\left(-4x^2+10\right)^3\right)\;.$
Now $f(x)=\dfrac1{-10x}$: it multiplies its input by $-10$ and then takes the reciprocal of that. Thus,
$f\left(g\big(h(x)\big)\right)=f\left(-5\left(-4x^2+10\right)^3\right)=\frac1{-10\left(-5\left(-4x^2+10\right)^3\right)}\;,$
which can be simplified to
$\frac1{50\left(-4x^2+10\right)^3}\;.$
You can even pull out a factor of $2^3=8$ if you wish:
$\frac1{50\left(-4x^2+10\right)^3}=\frac1{400\left(-2x^2+5\right)}\;.$