How can I show this field extension equality? How can I show this field extension equality   $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$?

The easier direction is $\mathbb{Q}(\sqrt{2} + \sqrt{3} ) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3}).$ Can you tell us why this is true? 
Now we have $\mathbb{Q} \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3} ) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3}).$ 

Since $\sqrt{2}$ is a root of the degree 2 polynomial $X^2-2$ which is irreducible over $\mathbb{Q}$ (because if it were reducible, it would have a linear factor corresponding to a rational root which the rational root theorem rules outs) we have $[\mathbb{Q}(\sqrt{2}): \mathbb{Q}]=2.$ 
Now $\sqrt{3}$ is a root of the degree 2 polynomial $X^2-3.$ Let's check this is irreducible over $\mathbb{Q}(\sqrt{2}).$ It is reducible if and only if it has a root in $\mathbb{Q}(\sqrt{2}).$ Again by the rational root theorem, it has no roots in $\mathbb{Q}.$ So if there is a root $a+b\sqrt{2} \in \mathbb{Q}(\sqrt{2})$ it must have $b\neq 0$ and satisfy $(a+b\sqrt{2})^2-3 = a^2+2ab\sqrt{2}+2b-3=0.$ If $a=0$ then $2b^2=3$ so $b=\pm \sqrt{3/2}$ which is rational if and only if $2b = \pm \sqrt{6}$ is rational. It is not hard to show that $\sqrt{n}$ is rational if and only if $n$ is a perfect square which $6$ is not, so $a+b\sqrt{2}$ is a root of $X^2-3$ then $a$ and $b$ must both be non-zero. But then we can rearrange to get $\sqrt{2} = \dfrac{3-2b+a^2}{2ab}$ and the right hand side is rational while the left is not. So $X^2-3$ is irreducible over $\mathbb{Q}(\sqrt{2})$ so $[\mathbb{Q}(\sqrt{2})(\sqrt{3}):\mathbb{Q}(\sqrt{2})]=2.$
Thus $$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2\cdot 2=4.$$

Now we find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}.$ The natural guess is $$(X-\sqrt{2}-\sqrt{3})(X-\sqrt{2}+\sqrt{3})(X+\sqrt{2}-\sqrt{3})(X+\sqrt{2}+\sqrt{3})=X^4+2X^2+25.$$ Let's check it is irreducible to confirm. Gauss's lemma allows us to simply check it is irreducible over $\mathbb{Z}.$ The rational root theorem gives $\pm 1, \pm 5, \pm 25$ as the only possibilities for integer roots, none of which work. So any factorization of $X^2+2X^2+25$ has no linear factors, and thus is of the form $$X^4+2X^2+25 = (X^2+aX+b)(X^2+cX+d)$$ where $a,b,c,d\in \mathbb{Z}.$ Equating coefficients of powers of $X$ gives $a=-c, ac+b+d=2, ad+bc=0$ and $bd=25.$ If $a=-c=0$ then $b+d=2$ and $bd=25$ which implies $4 = (b+d)^2 = b^2+2bd + d^2 = b^2+d^2+50 \implies b^2+d^2<0.$ Thus $a=-c\neq 0$ and from $ad+bc=0$ we get $b=d$ and from $bd=25$ we conclude $b=d=\pm 5.$ This makes $ac+b+d=2$ becomes $-a^2 = 2\pm 10$ which is impossible. 
Thus $$ [\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4.$$

Putting everything together, $$ 4 = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})][\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})]\cdot 4
$$
so $$[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}(\sqrt{2}+\sqrt{3})]=1$$ which implies $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3}).$
