The given function is $f(x)=x^{200}-x^{100}$, and I'm supposed to find it's decreasing and increasing intervals. Also, I should find them not by using derivatives but by doing function composition, like this:
$f_1(x)=x^{100}$
$f_2(x)=x(x-1)$
$f(x)= f_2(f_1(x))$
I know that $f_1$ is decreasing on the interval $(-\infty,0]$ and increasing on $[0,+\infty)$, and $f_2$ is decreasing on $(-\infty, \frac 12]$ and increasing on $[\frac 12, +\infty)$, but I'm not really sure what to do next.
First of all note that $f(x)$ is an even function, i.e. symmetric in $y$-axis. Thus it is enough to discuss the behaviour of $f(x)$ for $x\geq 0$. I use the same function $f_1(x)$ and $f_2(x)$ in the question.
For $x\leq y\leq \sqrt[100]{1/2}$ we have $f_1(x)\leq f_1(y) \leq 1/2$ so $f_2(f_1(x)) \geq f_2(f_1(y))$ hence $f(x)\geq f(y)$. So $f(x)$ is decreasing on $[0,\sqrt[100]{1/2}]$.
For $\sqrt[100]{1/2} \leq x\leq y$ we have $1/2\leq f_1(x)\leq f_1(y) $ so $f_2(f_1(x)) \leq f_2(f_1(y))$ hence $f(x)\leq f(y)$. So $f(x)$ is increasing on $[\sqrt[100]{1/2},\infty)$.
To keep the fun not only for myself, I let you do the work for $x<0$ (Hint: use symmetry).
Long comment:
Once you take (as you did) $f_1(x)=x^{100}$ and $f_2(x)=x^2-x$, you can find the $x$ coordinate of the vertex (here the minimum) of the parabola $f_2$, which is $x_0=-\frac{b}{2a}=\frac12$, so we need $x^{100}=\frac12\implies x_{1,2}=\pm\frac1{\sqrt[100]{2}}$.
Then you can factorise $f_2(f_1(x))=x^{100}\left(x^{100}-1\right)=x^{100}\left(x^{50}-1\right)\left(x^{50}+1\right)=x^{100}\left(x^{25}-1\right)(x^{25}+1)(x^{50}+1)$.
Roots of $f_2$ are $x\pm 1$ and $x=0$.
Now, investigate the behaviour of this product on the intervals $(-\infty,-1],(-1,0],(0,1]$ and $(1,+\infty)$.
We can leave $x^{100}$ and $x^{50}+1$ aside as those are never negative, their restrictions to $(-\infty,0)$ and $[0,+\infty)$ are separately injective and there is an even number of such factors.
$$\begin{array}{c}&&-\infty&&-1&&0&&1&^+\infty\\\\\hline x^{25}-1&&&-&|&-&|&-&\circ&+\\\hline x^{25}+1&&&-&\circ&+&|&+&|&+\\\hline \text{product}&&&+&&-&&-&&+\end{array}$$
Our $x_0=\pm \frac1{\sqrt[100]{2}}$ are exactly in the intervals $(-1,0]$ and $(0,1]$ respectively.
Now we have a better insight how $f(x)=x^{200}-x^{100}$ should look like $\left(-\infty,\frac12\right]\cup\left[\frac12,+\infty\right)$. Then we examine the intervals $\left(0,+\frac1{\sqrt[100]{2}}\right]$ and $\left(\frac1{\sqrt[100]{2}},+\infty\right)$ to see that the overall composition is decreasing on $\left(-\infty,-\frac1{\sqrt[100]{2}}\right)\cup\left(0,\frac1{\sqrt[100]{2}}\right)$ and increasing on $\left[-\frac1{\sqrt[100]{2}}, 0\right]\cup\left[\frac1{\sqrt[100]{2}},+\infty\right)$
