Take the following distributions :
X : -2, 0, 2 with
P(X) : 1/3, 1/3, 1/3

&

Y : -1, 1 with
P(Y) : 1/2, 1/2.

Now V(Z) = V(xy + x/y)
= V(xy) + V(x/y) + Cov(xy, x/y)
Now form distributions of xy and x/y and calculate their variances from that.
As for the covariance, it's coming as :
E((xy.x/y)^2) - (E(xy.x/y))^2 {standard formula of covariance).
It'll become : E((x^2)) - (E(x))^2
ie it's the variance of x only.
Both terms positive.
Therefore : greater.

Oh sorry I got it wrong by mistake. Yeah your formula is right but by that too the covariance is coming 8/3 (for the example distributions that I've taken) as the E(xy) term is 0.

We both have solved it taking example distributions only. You just made the distribution of y simpler with y=1 with probability 1. So your method is better. Thanks.