If the curve y=(f(x))/(f'(x)) tan lambda intersects

If the curve $y=\dfrac{f(x)}{f'(x)}\tan\lambda$ intersects the $Ox$ axis at several points, show that it always makes the same angle $\lambda$ with $Ox$ at each one.

Step 1: Where the curve meets $Ox$

Since $\tan\lambda\neq0$, $y=0$ forces $f(x_0)=0$: the curve meets $Ox$ exactly at the zeros of $f$.

Step 2: Derivative of $g(x)=\dfrac{f(x)}{f'(x)}\tan\lambda$

\[ g'(x) = \tan\lambda\cdot\frac{[f'(x)]^{2}-f(x)f''(x)}{[f'(x)]^{2}}. \tag{1} \]

Step 3: Slope at a zero $x_0$ of $f$

Since $f(x_0)=0$, the second term in the numerator of (1) vanishes: \[k = g'(x_0) = \tan\lambda\cdot\frac{[f'(x_0)]^{2}}{[f'(x_0)]^{2}} = \tan\lambda. \tag{2}\]

(This uses only $f'(x_0)\neq0$, i.e.\ that $x_0$ is a simple zero of $f$ — guaranteed at a genuine crossing of $Ox$.)