Z transform getting different answer for transform of rational function
Your first line is wrong. The partial fraction expansion of $X(z)$ should be :

$$ X(z) = \frac{ z (2z-\frac{5}{6})}{ (z-\frac{1}{2})(z-\frac{1}{3})} = 2 + \frac{ \frac{1}{2} }{z -  \frac{1}{2}} + \frac{ \frac{1}{3}}{z -  \frac{1}{3}} $$ 

Then they should yield the same inverse (causal) transform as:

$$ x[n] = 2\delta[n] + (1/2)^{n} u[n-1] + (1/3)^{n} u[n-1] $$

which is identical to :
$$ x[n] = (1/2)^n u[n] + (1/3)^n u[n] $$

as obtained by the second method.