As you say, four points (without their associated "$t$" values) are not enough to define a planar cubic Bezier curve.
In fact, 6 points are needed. But, on the other hand, the 6 points have to satisfy certain conditions, or else the required curve does not exist. In your case, you know that the curve exists, because it was used to generate the points in the first place.
The full story is given in this paper: J. Kozak, M. Krajnc, Geometric interpolation by planar cubic polynomial curves, Comput. Aided Geom. Des., 24 (2007), pp. 67-78. You can get a copy of the paper at the authors website.
Another approach (instead of the clever techniques suggested in the paper) is to use brute-force numerical methods. Start with 6 points $X_1, \ldots, X_6$ and 6 parameter values $t_1, \ldots, t_6$. We can assume that $t_1 = 0$ and $t_6 = 1$, but $t_2, t_3, t_4, t_5$ are unknown.We want to find control points $P_1, P_2, P_3, P_4$ for our curve. Obviously we can set $P_1 = X_1$ and $P_4 = X_6$, but $P_2$ and $P_3$ are unknown. We can (numerically) find values of $t_2, t_3, t_4, t_5$ and $P_2$ and $P_3$ that minimize the quantity
$$ \sum_{i=2}^{i=5}\Vert C(t_i) - X_i \Vert^2 $$
where $C(t)$ is the Bezier curve with control points $P_1, P_2, P_3, P_4$. This is a fairly nasty non-linear problem, but a good numerical algorithm will usually converge to the desired solution, given a decent starting point.
Another numerical approach: with the same notation we used above, solve the equations:
$$ C(t_i) = X_i \quad (i = 2,3,4,5)$$
Since the $X_i$ are 2D points, there are 8 equations here. We have 8 unknowns -- $t_2, t_3, t_4, t_5$ and the $x$ and $y$ coordinates of $P_2$ and $P_3$. The equations are non-linear, but a good numerical root-finding package should be able to handle them.
A much better approach is to get the Bezier curves from the CAD package. If the design guys won't give you the curves, you should charge them more money for your services, since they are making your life so much more difficult. Their choice -- curves or cash :-)
It depends what you're graphing.
The buttons on the left of your image say "xCurve, yCurve, zCurve", and this suggests that you have a 3D curve, and you are graphing one of the coordinates ($x$) versus a time parameter, $t$.
If so, the graph of the derivative is certainly wrong. It should have the value $0$ when the abscissa (the horizontal axis value, the $t$-value) is around 17 or 53.
On the other hand, your graphs don't look like "$x$ versus $t$" graphs, they look like "$(x,y)$ versus $t$" graphs. If this is the case, then your results might well be correct (though undesirable). See below for details.
Let's start from the beginning with a nice simple notation:
Suppose $P(t)$ is a cubic Bezier, with control points $A$, $B$, $C$, $D$. Then its equation is:
$$P(t) = (1-t)^3A + 3t(1-t)^2B + 3t^2(1-t)C + t^3D \quad (0 \le t \le 1) $$
Then the derivative curve is a quadratic (degree 2) curve, and its control points are $3(B-A)$, $3(C-B)$, $3(D-C)$, so it's equation is:
$$Q(t) = 3(1-t)^2(B-A) + 6t(1-t)(C-B) + 3t^2(D-C) \quad (0 \le t \le 1) $$
All of this applies regardless of whether $A$, $B$, $C$, $D$ are $x$ values or $(x,y)$ values.
If you want to draw "$x$ versus $t$" graphs, then drawing $P$ and $Q$ together on the same graph should be straightforward.
If you want to draw "$(x,y)$ versus $t$" graphs, then putting both $P$ and $Q$ on the same graph is more problematic. Suppose the control points $A$, $B$, $C$, $D$ were a great distance from the origin, but fairly close to each other. Then $B-A$, $C-B$, $D-C$ would be small, so the $Q$ curve would be close to the origin -- far away from the $P$ curve. In your case, it looks like (roughly) $A = (0,0)$ and $B=(32,12)$, so the first control point of the derivative curve $Q$ is $3(B-A) = (96,36)$, which is off the charts. Your derivative graph is clipped, so the end-points of the curve are not visible, which makes it harder to say whether or not it's correct. At least it looks like a parabola, though, which is correct (Bezier curves of degree 2 are parabolas).
These notes might help. Section 2.5 discusses derivatives, and there's a picture showing how the derivative curve relates to the original one (for the xy-vs-t case). Section 2.12 talks about the x-vs-t type of curve (which is variously referred to as a real-valued, explicit, or non-parametric Bezier curve).
Best Answer
When you say "gradient", I assume you mean the slope $dy/dx$.
First you get the derivative vector: $$ \left( \frac{dx}{dt}, \frac{dy}{dt} \right) $$ and then $$ \frac{dy}{dx} = \frac{\frac{dy}{dt} }{\frac{dx}{dt} } $$ As you might expect, this formula has problems when $dx/dt=0$, because this means you have a vertical tangent vector, so infinite slope.
From the general theory of Bezier curves, we know that the curve $$ f(t) = (1-t)^2 A + 2t(1-t) B + t^2 C $$ has derivative $$ \frac{df}{dt} = 2(1-t)(B-A) + 2t(C-B) $$ So, in your example $$ \frac{dX}{dt} = 2(1-t)(-1) + 2t(2) = 6t-2 $$ $$ \frac{dY}{dt} = 2(1-t)(-8) + 2t(2) = 20t - 16 $$ and so $$ \frac{dY}{dX} = \frac{\frac{dY}{dt} }{\frac{dX}{dt} } = \frac{20t-16}{6t-2} $$
Your reference to partial derivatives is confusing; partial derivatives make sense only when you have a function of several independent variables. In the case we're considering here, there is only a single variable, namely the parameter $t$.