There's no need to introduce work done here. The potential is related to the electric field strength by:
$\displaystyle \vec{E} =  \nabla V$
Assuming the scalar field $\displaystyle V(x,y,z) = c_0 c_1 x^3 + x  25 yz^4$, we can compute the gradient, $\displaystyle \nabla V$.
$\displaystyle \nabla V = \frac{\partial V}{\partial x} \hat{x} + \frac{\partial V}{\partial y} \hat{y} + \frac{\partial V}{\partial z} \hat{z}$
$\displaystyle = \left(3 c_0 c_1 x^2 +1\right) \hat{x}  25z^4 \hat{y}  100 y z^3 \hat{z}$
Assuming that the only force in the problem is the electric force, we have
$\displaystyle \vec{F} = \vec{E} q_0 = m_0 \vec{a}$.
Therefore,
$\displaystyle \vec{a} = \frac{\vec{E} q_0}{m_0} = \frac{q_0 \nabla V }{m_0}$
$\displaystyle = \frac{q_0}{m_0} \left(\left(3 c_0 c_1 x^2 +1\right) \hat{x} + 25z^4 \hat{y} + 100 y z^3 \hat{z}\right)$
You might want to doublecheck that the potential you wrote down is the same as the one I assumed...
