First of all, we can observe that
[tex]\sqrt{18} = \sqrt{2\cdot 9}=\sqrt{2}\cdot\sqrt{9}=3\sqrt{2}[/tex]
Then, we can rationalize the denominator by using the formula
[tex](a-b)(a+b)=a^2-b^2[/tex]
If we multiply both numerator and denominator by [/tex]\sqrt{8}+3[/tex] we have
[tex]\dfrac{3\sqrt{2}}{\sqrt{8}-3}\cdot\dfrac{\sqrt{8}+3}{\sqrt{8}+3}=\dfrac{3\sqrt{2}(\sqrt{8}+3)}{-1} = -3\sqrt{2}(\sqrt{8}+3)[/tex]
We can also observe that
[tex]\sqrt{8} = \sqrt{2\cdot 4}=\sqrt{2}\cdot\sqrt{4}=2\sqrt{2}[/tex]
And the expression simplifies into
[tex]-3\sqrt{2}(2\sqrt{2}+3) =-12-9\sqrt{2} = -3(4+3\sqrt{2})[/tex]