Given equation of parabola is $y^{2}=4 x$.
The equation of the normal to the parabola
$y^{2}=4 a x$ at $\left(a m^{2},-2 a m\right)$ is
$y=m x-2 a m-a m^{3}$
where $m$ is the slope of the normal.
Here, $a=1$, so the equation of the normal is
$y=m x-2 m-m^{3}$
Since, the normal is perpendicular to the line
$x+3 y+1=0$
Then, $m=$ slope of normal to line $x+3 y+1=0$ is
$ m \times\left(-\frac{1}{3}\right)=-1 $
$\Rightarrow m=3 $
$\therefore y=3 x-2(3)-(3)^{3} $
$\Rightarrow y=3 x-6-27 $
$\Rightarrow y=3 x-33$
$ \Rightarrow 3 x-y=33$