Yes, it's correct.
In fact, it would be correct, by definition, for any trajectory: if the path of an object in the $(x,y)$ plane can be written as $y = f(x)$ for some differentiable function $f$, and the object is moving along the path in the direction of increasing $x$, then the direction the object moves in (expressed as an angle from the horizontal) when it crosses the line $x = x_0$ is given by $\arctan f'(x_0)$.
As for why your results might not look like what you expect, well, the first question that comes to mind is whether your $\arctan$ function returns degrees or radians. Being off by a factor of $180/\pi$ would certainly make the results look odd.
If that's not the issue, then the next thing I'd check is whether you actually calculated $f'$ correctly.
Ps. I'm assuming that you did, in fact, mean $x$ to be horizontal position rather than time. However, if so, your definition of $f$ is somewhat odd. Generally, if a projectile is fired at time $t_0$ from position $(x_0,y_0)$ with speed $v_0$ and angle $\theta$, and is pulled down by gravity with acceleration $g$, the equations of motions are
$\begin{aligned} x &= x_0 + v_0 \cos(\theta)\, t \\ y &= y_0 + v_0 \sin(\theta)\, t - \frac12 g t^2, \end{aligned}$
the first of which we may solve for $t$ and substitute into the second to get the trajectory
$\begin{aligned} y &= y_0 + v_0 \sin(\theta) \frac{x-x_0}{v_0 \cos(\theta)} - \frac12 g \left(\frac{x-x_0}{v_0 \cos(\theta)}\right)^2 \\ &= y_0 + \tan(\theta) (x-x_0) - \frac{g}{2v_0^2\cos^2(\theta)}(x-x_0)^2. \end{aligned}$
Assuming that $x_0 = y_0 = 0$, and comparing the result with your original equation, we see that the coefficient $\tan(\theta)$ corresponds to your $V_0$, which is thus not a velocity but a slope, while the coefficient $-9.8$ (or $-4.9$ as you edited it) corresponds not to $-g$ or $-g/2$, but to $-g/(2v_0^2\cos^2(\theta))$. Only in the special case where the horizontal velocity component $v_0\cos(\theta) = 1$ are the latter two expressions equal (as they should be, given that in that case $t = x$).