Prove that $X\cdot (\nabla \times X)=0$, where $X \perp T_x$ 
Consider a $C^1$ map $f:R^2\to R^3$ such that $df_x:R^2\to R^3$ has rank $2$ everywhere. Define the tangent plane $T_x$ as $df_x(R^2)\subset R^3$. Suppose a vector field $X$ in $R^3$ is orthogonal to $T_x$ for all $x$, i.e., $X(f(x))\cdot Y=0$ for all $x\in R^2$ and $Y\in T_x$. Show that $X\cdot (\nabla \times X)=0$ at all points $f(x)$.

My thoughts: any vector field on $R^3$ can be written as $a(p,q,r)\partial_p+b(p,q,r)\partial q+c(p,q,r)\partial _r$ where $p,q,r$ are coordinates in $R^3$. So $$\nabla\times X=(\partial_qc\partial_r-\partial_rb\partial_q,\partial_ra\partial_p-\partial_pc\partial_r,\partial_pb\partial_q-\partial_qa\partial_p)$$ and 
$$X\cdot (\nabla \times X)=a\partial_p\partial_qc\partial_r-a\partial_p\partial_rb\partial_q+b\partial_q\partial_ra\partial_p-b\partial_q\partial_pc\partial_r+c\partial_r\partial_pb\partial_q-c\partial_r\partial_qa\partial _p$$
I need to show that the above expression applied to $f(x)$ is zero for all $x\in R^2$. Further, $$X(f(x))=(a\partial_pf(x),b\partial_qf(x),c\partial_rf(x))\\Y=(\partial_xf_1u+\partial_yf_1 v,\partial_xf_2u+\partial_yf_2 v,\partial_xf_3u+\partial_yf_3 v)$$ where $f=(f_1,f_2,f_3), u,v,\in R$. And we are given that $X(f(x))\cdot Y=0$. Even if I expand this, I don't know what to do next... 
 A: Differential forms for the win.
Think of the image of $f$ locally as an (embedded) surface $M$. Take a nonzero $1$-form $\omega$ on an open subset of $\Bbb R^3$ so that $M$ is an integral manifold of $\omega=0$. [For example, express $M$ as a level set of a smooth function $g$ and take $\omega = dg$.] The fact that $\omega=0$ is integrable tells you that $d\omega \wedge\omega = 0$. 
Under the usual dictionary between $1$-forms and vector fields in $\Bbb R^3$, if $X$ corresponds to $\omega$, $\text{curl}\, X = \nabla\times X$ corresponds to the $2$-form $d\omega$, and the dot product $X\cdot(\nabla\times X)$ corresponds to the $3$-form $\omega\wedge d\omega$. (It is easy to check, in coordinates, that if you have vector fields $X,Y$ corresponding to $1$-forms $\omega,\psi$, then $X\cdot Y = \star(\omega\wedge\star\psi)$.)
By the way, note that if we replace $X$ by any functional multiple $\phi X$, $(\phi X)\cdot(\nabla\times \phi X) = \phi^2 X\cdot(\nabla\times X) + \phi X\cdot (\nabla\phi\times X) = 0$, as well. From the differential forms viewpoint, we're looking at $(\phi\omega)\wedge d(\phi\omega) = \phi^2 \omega\wedge d\omega + \phi\omega\wedge d\phi\wedge\omega$, and $\omega\wedge d\phi\wedge\omega = d\phi\wedge\omega\wedge\omega = 0$, since $\omega$ is a $1$-form. 
EDIT: More interesting, if we replace $X$ by $X+Z$ for any vector field $Z$ vanishing on $M$, why does $(X+Z)\cdot\nabla\times(X+Z)$ still vanish on $M$? This boils down to seeing that $X\cdot\nabla\times Z = 0$ on $M$. This is true the component of the curl orthogonal to $M$ comes only from partial derivatives of components of $Z$ in directions tangent to $M$, and these vanish because $Z=0$ on $M$. [This, too, is easier to see using differential forms, since we don't need to worry about a particular coordinate system. Choose local coordinates $(x,y,z)$ so that $M = \{z=0\}$. Taking $\omega = dz + \eta$, where $\eta = P\,dx+Q\,dy+R\,dz$ vanishes identically on $M$, we want to see that $d\omega\wedge\omega$ still vanishes on $M$. Well, $d\eta\wedge dz = \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) dx\wedge dy\wedge dz$, and these partials vanish along $z=0$ since the functions are identically $0$ on $z=0$.]
If you are not familiar with differential forms and integrability, I would recommend proceeding as follows. Write $M$ as $g=0$ and take $X = \nabla g$. (Recall that gradients are orthogonal to level sets.) Then $\nabla\times X = 0$ and we're done. From my earlier comment, rescaling $X$ won't change the result of the computation. 
If you want to justify the assertion that $M$ is (locally) a level set of a smooth function, you need something like the implicit function theorem. It is locally a graph over one of the coordinate planes, say $z=\psi(x,y)$. Then take $g(x,y,z) = z-\psi(x,y)$.
