Vector field and normal of the field are both gradient fields Are there any general conditions to use to find a vector field f(x,y) that is a gradient field and f(-y,x) is also a gradient field.  It seems to me like if their second partial derivatives are zero then this is true or at least I haven't find an exception to that yet. 
 A: $\vec{f}(x,y)=(f_1(x,y),f_2(x,y))$ is a gradient field iff we have $f_1 = \frac{\partial g}{\partial x}, f_2 = \frac{\partial g}{\partial y}$ for some function $g$. Similarly, $\vec{f}(-y,x)=(f_1(-y,x),f_2(-y,x))$ is a gradient field iff we have 
$$f_1(-y,x) = \frac{\partial h}{\partial x}(x,y) = -\frac{\partial h}{\partial y}(-y,x), f_2(-y,x) = \frac{\partial h}{\partial y}(x,y)=\frac{\partial h}{\partial x}(-y,x)$$
 for some function $h$. The second inequalities in each case are simply applications of the rules of differentiation. This means that $f$ is such that $f(x,y)$ and $f(-y,x)$ are both gradient fields iff
$$f_1 = \frac{\partial g}{\partial x} = -\frac{\partial h}{\partial y}$$
$$f_2 = \frac{\partial g}{\partial y} = \frac{\partial h}{\partial x}$$
can be solved by two functions $g,h$. If we let $h(x,y) = g(-y,x)$, we see that
$$\frac{\partial g}{\partial x} = -\frac{\partial h}{\partial y}$$
$$\frac{\partial g}{\partial y} = \frac{\partial h}{\partial x}$$
and so if $\vec{f}(x,y)$ is a vector field then $\vec{f}(-y,x)$ must be (and vice-versa).
A: If I understand the question correctly this is a somewhat unfamiliar setup. Let's see what comes out of it!
We are given a vector field
$${\bf v}(x,y):=\bigl(P(x,y), Q(x,y)\bigr)$$
defined in a neighborhood of $(0,0)\in{\mathbb R}^2$. In addition we consider the second vector field
$$\bar {\bf v}(x,y):=\bigl(\bar P(x,y),\bar Q(x,y)\bigr)$$
related to ${\bf v}$ via
$$\bar {\bf v}(x,y):={\bf v}(-y,x), \quad{\rm i.e.,}\quad
\bigl(\bar P(x,y),\bar Q(x,y)\bigr):=\bigl(P(-y,x),Q(-y,x)\bigr)\ .$$
Geometrically the field $\bar{\bf v}$ is obtained from the field ${\bf v}$ by attaching at $(x,y)$ the vector ${\bf v}(-y,x)$. The point $(-y,x)$ is obtained from $(x,y)$ by a $90^\circ$ rotation, but the vector found there is not turned.  This will have an important effect!
Now we are told that both fields ${\bf v}$ and $\bar {\bf v}$ are gradient fields. Writing $f_1$ (resp. $f_2$) for the partial derivative of some function $f$  with respect to the first (resp. second) variable we therefore have
$$P_2(x,y)\equiv Q_1(x,y)\ , \qquad  \bar P_2(x,y)\equiv \bar Q_1(x,y)\ .$$
Here $\bar P_2(x,y)$ and $\bar Q_1(x,y)$ compute as follows:
$$\eqalign{
 \bar P_2(x,y)&={\partial \over\partial y}P(-y,x)=-P_1(-y,x)\ , \cr
\bar Q_1(x,y)&={\partial \over\partial x}Q(-y,x)=Q_2(-y,x)\ . \cr}$$
This shows that we have $-P_1(-y,x)=Q_2(-y,x)$ identically in $x$ and $y$, and this is the same thing as $-P_1(x,y)=Q_2(x,y)$ identically in $x$ and $y$.
We now leave the field $\bar{\bf v}$ aside and write again $f_x$, $f_y$ for the partial derivatives. Then altogether we can say that the components $P$ and $Q$ of the original field ${\bf v}$ together satisfy the Cauchy-Riemann equations (up to a sign):
$$P_y=Q_x\ ,\qquad P_x=-Q_y\ .$$
It follows that the complex function $f(x,y):=P(x,y)-i Q(x,y)$ is an analytic function of the complex variable $z=x+ iy$.
