Y=f(x) is a differentiable function on R. When x≠0:

f '(x)+f(x)/x & gt； 0

Namely: [xf' (x)+f (x)]/x > 0

So: [xf (x)]'/x > 0

So:

x & lt0，[xf (x)]'

x & gt0，[xf(x)]' & gt； 0, g(x)=xf(x) is a monotonically increasing function.

So: g(x)=xf(x) gets the minimum value g(0)=0 when x=0.

So: g(x)>=0 holds.

F (x) = xf (x)+1/x = g (x)+1/x = 0, that is, g (x) =-1/x.

The inverse proportional function h(x)=- 1/x is a monotonically increasing function in the second quadrant and the fourth quadrant.

x & gt0，h (x)

X<0, h(x)>0, g(x)>=0, and h(x) and g(x) have a unique intersection.

To sum up, g(x)=- 1/x has a unique solution.

So: F(x)=xf(x)+ 1/x has a unique zero.

Option b