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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.

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The normal of a vector field $(f(x,y), g(x,y))$ is $(-g(x,y), f(x,y))$, not $(f(-y,x), g(-y,x))$ as you have written. – Rahul Jan 6 '12 at 4:09
Indeed, your question and its title ask different questions! – Mariano Suárez-Alvarez Jan 6 '12 at 4:22

$\vec{f}(x,y)=(f_1(x,y),f_2(x,y))$ 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).

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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 $z=x+ iy$.

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