slope - Can you resolve this contradiction concerning the geometric interpretation of differential equations? - Mathematics Stack Exchange

in 1st lecture of mit ocw differential equations course,

at around 25th minute, example y’=-x/y visualized multiple isoclines, intersect @ (0,0). @ (0,0) has multiple (possible?) slopes.

also if substitute (0,0) y'=-x/y or rather y*y'=-x, 0*y’=0, true slope y’.

but @ 37:45, professor argues "you can’t have 2 slopes" [at same point].

can resolve contradiction?

an ordinary differential equation has form $$y'=f(x,y)\qquad\bigl((x,y)\in\omega\bigr)\ ,$$ whereby given function $f$ defined , continuous in region, i.e. connected open set $\omega\subset{\mathbb r}^2$. if $f$is given "analytical expression" in variables $x$ , $y$, in example, tacit understanding should take $\omega$ maximal region $f$ defined , continuous. $f(x,y):=-{x\over y}$ undefined when $y=0$. therefore can take $\omega$ either upper or lower half plane of ${\mathbb r}^2$. both choices $f$ defined , real analytic on $\omega$. if take $\omega$ upper half plane solution curves of ode $y'=-{x\over y}$ upper semicircles $$\gamma_c:\quad y=\sqrt{c^2-x^2}\qquad (-c<x<c)$$ $c>0$.