I'm working with field extensions (all with characteristic 0) and I need to show that:
$|\mathbb{Q}(a):\mathbb{Q}|=n,|\mathbb{Q}(b):\mathbb{Q}| =m$ and $\gcd(m,n)=1 \implies|\mathbb{Q}(ab):\mathbb{Q}|=mn$
Here's what I thought: We know that $\mathbb{Q}(a,b) = \mathbb{Q}(ab,a) = \mathbb{Q}(ab,b)$, and $|\mathbb{Q}(a,b):\mathbb{Q}|=mn$
so, $$|\mathbb{Q}(a,b):\mathbb{Q}| = |\mathbb{Q}(ab,b):\mathbb{Q}| = |\mathbb{Q}(ab)(a):\mathbb{Q}(ab)|\cdot |\mathbb{Q}(ab):\mathbb{Q}|= n\cdot m $$
$\implies 1 = \frac{|\mathbb{Q}(a,b):\mathbb{Q}|}{|\mathbb{Q}(ab,a):\mathbb{Q}|} = \frac{n\cdot m}{|\mathbb{Q}(ab,a):\mathbb{Q}(ab)|\cdot |\mathbb{Q}(ab):\mathbb{Q}|}\implies |Q(ab):\mathbb{Q}| = \frac{n\cdot m}{|\mathbb{Q}(ab,a):\mathbb{Q}(ab)|}$
Now, as $$|\mathbb{Q}(a):\mathbb{Q}|=n \implies \deg(\operatorname{irr}(a,\mathbb{Q}))=n$$ we have that $$\deg(\operatorname{irr}(a,\mathbb{Q}(ab)))\leq n\implies |\mathbb{Q}(ab,a):\mathbb{Q}(ab)|\leq n$$
Therefore $|\mathbb{Q}(ab):\mathbb{Q}| = m\cdot \frac{n}{|\mathbb{Q}(ab,a):\mathbb{Q}(ab)|}= m \cdot k$, with $k \in \mathbb{Z}$.
By the same logic $|\mathbb{Q}(ab):\mathbb{Q}| = n \cdot t$, with $t \in \mathbb{Z}$
So $|\mathbb{Q}(a,b):\mathbb{Q}|$ is a multiple of $m$ and $n$, but as they are coprimes, the least commom multiple of them is their product, which is $mn = |\mathbb{Q}(a,b):\mathbb{Q}|$, and as $\mathbb{Q}(ab) \subset \mathbb{Q}(a,b), |\mathbb{Q}(ab):\mathbb{Q}|\leq|\mathbb{Q}(a,b):\mathbb{Q}|$, so we have that $|\mathbb{Q}(ab):\mathbb{Q}|=mn$ $\blacksquare$
Are there any mistakes?
As @Batominovski points out in a comment, the result that you are trying to prove is not true.
To answer your question about there being a problem with your proof, there is one: while it is true that $[\mathbb{Q}(ab)(a):\mathbb{Q}(ab)] \leq n$, it is not necessarily true that $[\mathbb{Q}(ab)(a):\mathbb{Q}(ab)]$ divides $n$.
This can be seen in Batominovski's example. Here $a=\sqrt[3]{2}$ and $b=\exp(\frac{2\pi i}{3})$, so $n=3$ and $m=2$. You rightly claim that $[\mathbb{Q}(\sqrt[3]{2} \exp(\frac{2\pi i}{3}))(\sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2} \exp(\frac{2\pi i}{3}))] \leq n=3$, but in this case $[\mathbb{Q}(\sqrt[3]{2} \exp(\frac{2\pi i}{3}))(\sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2} \exp(\frac{2\pi i}{3}))] = 2$, which does not divide 3.