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If I understand you correctly, you meant that $U_x=\frac{\partial U}{\partial x}$, right ? If so, your result is a triviality, since :

• There exists some quantity $K$ such as $U(x,y)=K x^\alpha$, where $K$ doesn't depend on $x$ (in fact, it is obvious that $K=A y^\beta$).

• Therefore $\frac{\partial U}{\partial x}=K\frac{\partial x^\alpha}{\partial x}=K\alpha x^{\alpha-1}$

This can be checked in sage quite directly:

sage: U(x,y)=A*x^a*y^b;U
(x, y) |--> A*x^a*y^b
sage: bool(U(x,y).diff(x)==a*U(x,y)/x)
True