Which sphere is fastest?

For X or Z: Final speed would be:$\sqrt{v^2+2gH}$(Using equation of motion for a uniformly accelerated motion). For Y , you saw, the net speed would be the resultant of horizontal and vertical components: $\sqrt{v^2+(\sqrt{2gH})^2}$.(horizontal component is v, and vertical is $\sqrt{2gH}$. So, kinematics and energy conservation don't contradict! Good efforts, though.

Note: you would have observed, how energy conservation is more convenient to use in such situations. Makes life easier!

Your logic seems to be the following:

Suppose that the velocity that a ball would reach if it were just dropped is $v_1$. Suppose that it's instead thrown with velocity $v_2$. The velocity of the horizontally thrown ball will be $\sqrt{v_1^2+v_2^2}$, but the velocity of the vertically thrown ball will be $v_1+v_2$. Since $v_1+v_2 \neq \sqrt{v_1^2+v_2^2}$, we have a contradiction.

There are a few problems with that.

One is that you didn't explicitly state your claim that the vertically thrown ball will have velocity $v_1+v_2$. It's important to make your reasoning as explicit as possible, not only so other people can follow, but also because it makes it more likely that you yourself will spot your error.

The other problem is that this claim is wrong. There are several ways of seeing this:

-Momentum equation Impulse, or change in momentum, is equal to $force \times time$. The ball thrown downwards will be in the air less time than the dropped ball, so the gravitational force will be acting for less time. Thus, the impulse imparted by gravity to the thrown ball will be less than the impulse imparted to the dropped ball, and thus the change in velocity will be smaller.

-Conservation of energy The dropped ball starts out with an energy of $PE = mgh$, and ends with an energy of $KE = \frac 12 mv_1^2$. So $mgh = \frac 12 mv_1^2$. The thrown ball starts out with an energy of $KE+PE = \frac 12 mv_2^2+mgh$. We can then substitute $mgh = \frac 12 mv_1^2$ into that equation and get the the thrown ball starts out withh $KE+PE = \frac 12 mv_2^2+\frac 12 mv_1^2$. If its final velocity is $v_3$, then by conservation of energy we have $\frac 12 mv_3^2 = \frac 12 mv_2^2+\frac 12 mv_1^2$. Canceling out the $\frac12 m$, we get that $v_3^2 = \ v_2^2+v_1^2$, or $v_3= \sqrt{v_1^2+v_2^2}$, which is the same as for the horizontally thrown ball.

-Kinematics equations You should be familiar with the formula that for constant acceleration, $v_{final}^2-v_{initial}^2=2ad$. This can be derived from conservation of energy similar as above, or by using calculus. This gives $v_{final}^2=v_{initial}^2+other stuff$ rather than $v_{final}=v_{initial}+other stuff$, as you seem to be assuming.

So, the takeaway is that often, it's not velocities, but the squares of velocities, that add.