Clairaut's Equation Singular and General Solutions

First note that your general solution is not entirely correct (you can check it by plugging it into your equation). The correct solution is given by $$ y_g=Cx-e^C $$ and depends on only one arbitrary constant.

Your singular solution is correct and given by $$ y_s=x\ln x-x. $$ How to show that actually there is always a point at which the general and singular solutions are tangent? Consider the condition that two functions are tangent at a point $x_0$: $$ y_g(x_0)=y_s(x_0),\quad y'_g(x_0)=y'_s(x_0).\tag{1} $$ (They should have the same $y$ coordinates and the same slope, and this is exactly what is written there.) From you solutions and using the second equality in $(1)$ you find that $$ C=\ln x_0. $$ Putting this value into the equality $y_g(x_0)=y_s(x_0)$ you will find an identity which proves that your singular solution is tangent to the general solutions.

Here is the picture: