#### Answer

The particular solution that passes through the point $(3,4)$ is
$$y=\sqrt{2x^2-2}.$$

#### Work Step by Step

Here the solution is given in its' implicit form $2x^2-y^2=C$ and it contains an arbitrary constant $C$. To find required $C$ we have to demand that the solution passes through the point $(3,4)$ i.e. when we put $x=3$ we get $y=4$:
$$2\times3^2-4^2=C\Rightarrow C=2.$$
Now we have
$$2x^2-y^2=2.$$
From here we will find $y$ explicitly:
$$2x^2-y^2=2\Rightarrow y^2=2x^2-2\Rightarrow y=\pm\sqrt{2x^2-2}.$$
We have to throw away the solution with the $"-"$ sign because when we put $x=3$ we get
$$y=-\sqrt{2\times3^2-2}=-\sqrt{16}=-4$$
which we did not want.
When we take the $"+"$ sign we have for $x=3$
$$y=\sqrt{2\times3^2-2}=\sqrt{16}=4,$$
and that is exactly what we wanted.
So the particular solution that passes through the point $(3,4)$ is
$$y=\sqrt{2x^2-2}.$$