Is the composition of irreducible polynomials again irreducible? [closed]

This need not be true. Note that in $\mathbb{R}[x]$ any irreducible polynomial has degree either $1$ or $2$. So, you can take two irreducible polynomials of degree 2, and compose them to get a reducible polynomial.

I am extending @Pipicito's answer for the case when both polynomials have degree greater than 1. The answer is also no. For example, consider $f(x)=x^2-\frac{4}{3}$; it is irreducible over $\mathbb{Q}$. However, $f(f(x))={(x^2-\frac{4}{3})}^2-\frac{4}{3}=(x^2-2x+\frac{2}{3})(x^2+2x+\frac{2}{3})$ and it is reducible!

Moreover, there is Cappelli's lemma about this question. It says the following:

Let $K$ be a field, $f$ and $g$ be irreducible polynomials over $K$. Then, $f(g(x))$ is irreducible over $K$ iff $f$ is irreducible over $K$ and $g-\alpha$ is irreducible over $K(\alpha)$ for every root $\alpha$ of $f$.

In the example, the roots of $x^2-\frac{3}{4}$ are $\pm \frac{\sqrt{3}}{2}$. So, to show $f(f(x))$ is reducible, just show that one of $x^2-\frac{3}{4}\pm \frac{\sqrt{3}}{2}$ has roots in $\mathbb{Q}(\sqrt{3})$