The curve $x^2–2xy+y^2+3x–3y = 0$ cuts the x-axis at (0, 0) at an angle

The given curve is $x^2–2xy+y^2+3x–3y = 0$

Use implicit differentiation to find $\frac{dy}{dx}$

$2x-2x\frac{dy}{dx}-2y+3-3\frac{dy}{dx}=0$

$\begin{array}{rcl}\frac{dy}{dx}\left(2x+3\right)& =& 2x-2y+3\\ \frac{dy}{dx}& =& \frac{2x-2y+3}{2x+3}\end{array}$

Let $m$ be the gradiant of the curve at origin, so that $m=\frac{dy}{dx}at \left(0,0\right)$

$m=1$

Since the slope of the tangent to the curve at origin is 1, the inclination is ${\tan }^{-1}1=\boxed{\frac{\mathrm{\pi }}{4}}$