Prove that if a particle moves in constant speed, its acceleration is orthogonal to its velocity. I need help with this problem:

Prove that if a particule moves in constant speed, then its acceleration is orthogonal to its velocity.

I tried to prove it like this:
If the speed of $f$ is constant, that means that $\Vert f'(t)\Vert=\Vert(x'(t), y'(t))\Vert=c$. If the acceleration is orthogonal to the velocity, that means that $f'(t)\cdot f''(t)=0$, thus $x'(t)x''(t)+y'(t)y''(t)=0$ and that means that $x'(t)x''(t)= -y'(t)y''(t)$.
I really think I'm wrong. Please can you help me prove this correctly?
 A: If the speed $\Vert f'(t)\Vert$ is constant then its derivative is zero. The calculation becomes even simpler if we consider the square of the speed (which is constant as well):
$$
 0 = \frac{d}{dt} \Vert f'(t)\Vert^2 = \frac{d}{dt} \left( x'(t)^2 + y'(t)^2 \right)
= 2  x'(t) x''(t) + 2 y'(t) y''(t) = 2 f'(t) \cdot f''(t)
$$
so that $f'(t) \cdot f''(t) = 0$, i.e. the speed and the acceleration are orthogonal at each time.
A: Let $\vec v(t)$ be the velocity vector.
$\vec v \cdot \vec v = v^2 = C$.
Differentiate the dot product:
$2\dfrac{d(\vec v)}{dt} \cdot \vec v=0;$
$\vec a \cdot \vec v =0.$
Acceleration vector $ \vec a$ is perpendicular to velocity vector $\vec v$.
A: \begin{align}
& \color{red}{\large 0} =
{\mathrm{dv}\left(t\right) \over \mathrm{d}t} =
{1 \over 2\mathrm{v}\left(t\right)}{\mathrm{dv^{2}}\left(t\right) \over \mathrm{d}t} =
{1 \over 2\mathrm{v}\left(t\right)}
{\mathrm{d}\left[\vec{\mathrm{v}}\left(t\right)\cdot\vec{\mathrm{v}}\left(t\right)\right] \over \mathrm{d}t} =
{\vec{\mathrm{v}}\left(t\right) \over \mathrm{v}\left(t\right)}
\cdot
{\mathrm{d}\vec{\mathrm{v}}\left(t\right) \over \mathrm{d}t}
\\[5mm] &\
\implies \bbox[10px,border:1px groove navy]{\vec{\mathrm{v}}\left(t\right) \perp \vec{\mathrm{a}}\left(t\right)}\,,\qquad
\vec{\mathrm{a}}\left(t\right) \equiv
{\mathrm{d}\vec{\mathrm{v}}\left(t\right) \over \mathrm{d}t} 
\end{align}
