Equation of Normal Line

Given curve,

$$y=x\log_ex$$

Differentiate with respect to $$x$$: $$\frac{dy}{dx}=1+\log_ex$$

The line $$2x-2y=3$$ can be written as $$y=x-\frac32$$

Hence its slope is $$1$$

Since the normal is parallel to this line, the slope of the normal is $$1$$

Therefore, the slope of the tangent is $$-1$$

Thus,

$$1+\log_ex=-1$$

$$\log_ex=-2$$

$$x=e^{-2}$$

Now substitute into the curve:

$$y=x\log_ex$$

$$=e^{-2}(-2)$$

$$=-2e^{-2}$$

Hence, the required point is

$$\boxed{\left(e^{-2},-2e^{-2}\right)}$$