Is $ds$ and curvature has the max value at the same point? If there is a plane curve, define as a $y=f(x)$, and my guess is if curvature of this curve $\frac{y^{\prime\prime}}{(1+(y\prime^2))^\frac{3}{2}}$ has a max value at ${x=x_0}$ and $\sqrt{dx^2+dy^2}$ has the max value at same point. that two is necessity, or sufficiency or both of them, or not anything? I try to derive them and both have $0$at same point, but this method not worked well.
I assume you are using the parametrization $r(t)=(t,f(t))$

Then the maximum of the speed is at

$$\frac{d^2 s}{dt^2}=\frac{f'(t)f''(t)}{\sqrt{1+f'(t)^2}}=0$$

Which would imply that $f'(t)=0$ or $f''(t)=0$, but we reject the case for $f'(t)=0$ because curvature cannot be defined on non regular curve. 

So that means the speed is at maximum when the second derivative of the curve is $0$.

$$\kappa '(t)=\frac{f'''(t)(1+f'(t)^2)^{\frac 32} - \frac 32 (1+f'(t)^2)^{\frac 12} (2f'(t)f''(t))f''(t)}{(1+f'(t)^2)^3}$$

It is equal to zero when the numerator is $0$ :

$$f'''(t)(1+f'(t)^2)^2 -3f'(t)f''(t)^2=0$$

If $f''(t)=0$ , we must have $f'''(t)=0$ also in order for the curvature to be at maximum, so we cannot conclude that.

In general, the curvature of a plane curve is an intrinsic property in the sense that it is independent of the parametrization and rigid transformation. So usually there would be no relation between the curvature and the derivative of the curve.
I assume you are using the parametrization $r(t)=(t,f(t))$

Then the maximum of the speed is at

$$\frac{d^2 s}{dt^2}=\frac{f'(t)f''(t)}{\sqrt{1+f'(t)^2}}=0$$

Which would imply that $f'(t)=0$ or $f''(t)=0$, but we reject the case for $f'(t)=0$ because curvature cannot be defined on non regular curve. 

So that means the speed is at maximum when the second derivative of the curve is $0$.

$$\kappa '(t)=\frac{f'''(t)(1+f'(t)^2)^{\frac 32} - \frac 32 (1+f'(t)^2)^{\frac 12} (2f'(t)f''(t))f''(t)}{(1+f'(t)^2)^3}$$

It is equal to zero when the numerator is $0$ :

$$f'''(t)(1+f'(t)^2)^2 -3f'(t)f''(t)^2=0$$

If $f''(t)=0$ , we must have $f'''(t)=0$ also in order for the curvature to be at maximum, so we cannot conclude that.

In general, the curvature of a plane curve is an intrinsic property in the sense that it is independent of the parametrization and rigid transformation. So usually there would be no relation between the curvature and the derivative of the curve.