If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$. If $\gcd(a,n) = 1$ and $\gcd(b,n) = 1$, then $\gcd(ab,n) = 1$.
Also... $a,b$ and $n$ are natural numbers. 
I feel I should begin with EEA to multiply out the gcd's, but I don't know where to go from there...
Because $n$ is coprime to $a$ and $b$, we know there are linear combinations that equal one. Let the coefficients be $x,y,z,w$ such that $ax + ny = 1$ and $bz + nw = 1$. Multiply these together:
$$abxz + axnw + bzny + n^2yw = 1 \cdot 1$$
$$ab(xz) + n(axw + bzy + nyw) = 1$$

So there's a linear combination of $ab$ and $n$ that is equal to $1$. Thus they are coprime.
There is a useful theorem for these kind of problems that says:

For $a,b \in \Bbb Z,a\neq 0, b\neq 0$ then $gcd(a,b)=1 \Leftrightarrow \exists m,n \in \Bbb Z : am+bn=1$

wich is not difficult to prove.

It is a useful exercise to prove or disprove this theorem when $gcd(a,b)=d$ and $am+bn=d$ where $d\ge 1$