Answer
The vector field $F$ is irrotational.
Work Step by Step
The vector field $F$ will be irrotational when $curl F=0$
Let us consider that $F=P i+Q j+R k$
Then, we have $curl F=[R_y-Q_z]i+[P_z-R_z]j+[Q_x-P_y]k$
As we are given that $F(x,y,z)=f(x) i+g(y) j+h(z) k$
Here, we have $curl F= \nabla \times F=[R_y-Q_z]i+[P_z-R_z]j+[Q_x-P_y]k$
or, $=(\dfrac{\partial h(z)}{\partial y}-\dfrac{\partial g(y)}{\partial z})i+(\dfrac{\partial f(x)}{\partial z}-\dfrac{\partial h(z)}{\partial x})j+(\dfrac{\partial g(y)}{\partial x}-\dfrac{\partial f(x)}{\partial y})k=0$
Hence, we have: the vector field $F$ is irrotational.
