Determine the point(s) on the curve $y=\frac{x^{2}}{\ln x}$ where the tangent is horizontal.

Step 1 ：Find the derivative of the function $y = \frac{x^2}{\ln x}$: $\frac{dy}{dx} = \frac{2x}{\ln x} - \frac{x}{(\ln x)^2}$

Step 2 ：Set the derivative equal to 0: $\frac{2x}{\ln x} - \frac{x}{(\ln x)^2} = 0$

Step 3 ：Solve the equation for x: $x = e^{\frac{1}{2}}$

Step 4 ：Final Answer: The point(s) on the curve $y=\frac{x^{2}}{\ln x}$ where the tangent is horizontal is at $x = \boxed{e^{\frac{1}{2}}}$