The easiest way to determine the degree of the extension (that is, the dimension of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$) is to use the multiplicativity of the degree:
$$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}].$$
We know that $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2$. Since $\sqrt{3}$ satisfies $x^2-3\in\mathbb{Q}(\sqrt{2})[x]$, this means that $[\mathbb{Q}(\sqrt{2})(\sqrt{3}):\mathbb{Q}(\sqrt{2})]\leq 2$, and so is equal to either $1$ or $2$. So really the only question is whether this extension has degree $1$ or $2$.
The extension has degree $1$ if and only if $\sqrt{3}\in\mathbb{Q}(\sqrt{2})$. This amounts to checking if there exists $a,b\in\mathbb{Q}$ such that $(a+b\sqrt{2})^2 = 3$. It is straighforwad to determine the answer with basic algebra. And so this completely gives you the answer.
Trying to show that $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ is linearly independent over $\mathbb{Q}$ is much more time consuming, but amounts to essentially the same thing. The equation you got,
$$a+2b^2-3c^2-6d^2=(6cd-2ab)\sqrt{2}$$
tells you that you must have $6cd-2ab=0$ and $a+2b^2-3c^2-6d^2=0$, because the left hand side is a rational, while the right hand side would be irrational if $6cd-2ab\neq 0$. So you know that $ab=3cd$, and that should let you proceed through a somewhat laborious and probably annoying process to conclude that you must have $a=b=c=d=0$.
