The graph of the equation $x^2 + y^2 = 1$ gives a circle of radius $1$, centered at the origin:
After we find its derivative:
$$ \frac{dy}{dx} (x^2 + y^2) = \frac{dy}{dx} 1 \\
2x + 2y \frac{dy}{dx} = 0 \\
\frac{dy}{dx} = -\frac{x}{y}
$$
if we graph it on a 3D space we get this shape:
and it does describe the slope of the circle (positive in quadrant II & IV, negative in I & III)
Now let's consider a "similar" equation: $$ x^3 + xy^2 = x $$ The graph should be same everywhere except when $x = 0$, since it's just the last equation multiplied by $x$ on both sides, and the differentiability of it should be the same except at $x = 0$
However, if we work out the derivative of y with respect to x:
$$ \frac{dy}{dx} (x^3 + xy^2) = \frac{dy}{dx} x \\
3x^2 + y^2 + 2xy \frac{dy}{dx} = 1 \\
\frac{dy}{dx} = -\frac{3x^2+y^2-1}{2xy}
$$
we get a completely different function, and plotting the graph yields this weird shape:
and it doesn't describe the slope of the circle at all. So why does finding the derivative of the first equation works, but not the second?
The only difference between the two derivatives is entirely due to the introduction of the vertical line $ \ x = 0 \ $ that is combined with the circle. The derivative for the second "curve" is $$ \frac{dy}{dx} \ \ = \ \ \frac{1 \ - \ 3x^2 \ - \ y^2 }{2xy} \ \ = \ \ \frac{(\overbrace{1 \ - \ x^2 \ - \ y^2}^{= \ 0}) \ - \ 2x^2 }{2xy} \ \ = \ \ -\frac{ x^2 }{ xy} \ \ . $$
For points on the circle that are not on the $ \ y-$axis, both derivative expressions produce the same result: for example, at $ \ ( -\frac45 \ , \ \frac35 ) \ , $ we have $$ \frac{dy}{dx} \ \ = \ \ -\frac{-\frac45}{\frac35} \ \ = \ \ \frac43 \ \ \ \ \text{versus} \ \ \ \frac{dy}{dx} \ \ = \ \ \frac{1 \ - \ 3·\frac{16}{25} \ - \ \frac{9}{25} }{2·\left(-\frac45 \right) ·\frac35} \ \ = \ \ -\frac{\frac{ - 32}{25} }{\frac{24}{25}} \ \ = \ \ \frac43 \ \ . $$
For the circle, we simply find horizontal tangents at $ \ (0 \ , \ \pm 1) \ \ , $ while $ \ x^3 + xy^2 \ = \ x \ \ $ has an indeterminate "value" $ \ -\frac{ 0^2 }{ 0·y} \ $ at all the points on the $ \ y-$axis [it is ambiguous at $ \ (0 \ , \ \pm 1) \ \ $ because there are both horizontal and vertical tangents at those points].