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Lets say that we have a parametric surface of the form $S(u, v) = [x, y, z]^\top$ where $S :\rm I\!R ^2\rightarrow \rm I\!R^3$.

Provided that for a point $\rm p \in I\!R ^2$ that lies on the surface $S$, we can compute its global position (i.e., $\rm{S}(p) = [x_p, y_p, z_p]^\top$), the first derivatives (i.e, $\rm{S}_u(\rm p)$ and $\rm{S}_v(\rm p)$), the second derivatives (i.e, $\rm{S}_{uu}(\rm p)$ and $\rm{S}_{vv}(\rm p)$) and the normal vector of the surface at $\rm p$ (i.e., $\rm N_s(p)$).

Given this information can we compute the mixed derivative of the surface $\rm S$ at $\rm p$ (i.e., $\rm{S}_{uv}(\rm p) = \rm{S}_{vu}(\rm p)$)?

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    that depends on the complexity of the functions $x,y,z$ – 2017-02-23
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    @janmarqz $x, y, z$ are not functions. $[x, y, z]^\top$ is a point in the 3D Eucledian space. – 2017-02-23
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    they oughta be, cuz the (x,y,z)^t form a surface in IR^3 – 2017-02-23
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    @janmarqz Let's say that they're Nurbs or Bezier. – 2017-02-23
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    In general, the answer is: NO. Reason is that the normal can be expressed in the first derivatives alone. And the second order derivatives contain no information about the mixed derivative. – 2017-03-04

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Let $S(u,v)=(u,v, kuv)$, denote a one parameter family of hyperbolic paraboloids. Note that $S_u(0,0)=(1,0,0), S_v(0,0)=(0,1,0), S_{uu}(0,0,0)=(0,0,0)=S_{v,v}(0,0)$, $N(0,0)=(0,0,1)$ : these quantities do not depend on $k$. But $S_{u,v}(0,0)=k$ does.