You can apply the product rule when differentiating $\vec v\cdot\vec v =$constant.
The intuitive idea is that $\vec v(t)$ traces out a curve on a sphere centered at the origin (i.e., picture $\vec v(t)$ as a moving radius vector), while $\vec v'(t)=\vec a(t)$ is tangent to the sphere, and hence perpendicular to the radius at the point of tangency, namely $\vec v(t)$.
There is also a geometric description of what this says in terms of the original curve, say $\vec x(t)$, of which $\vec v(t)$ gives the velocity. Since $\vec v(t)$ gives a vector tangent to the trajectory of $\vec x(t)$, and in your case $\vec a(t)$ is perpendicular to $\vec v(t)$, $\vec a(t)$ is perpendicular to the trajectory of the original curve $\vec x(t)$.
Acceleration can occur for 2 reasons. In general, the component of $\vec a(t)$ in the direction of the trajectory (or opposite this direction) tells you how the speed is changing, while the component of $\vec a(t)$ perpendicular to the trajectory tells you how the direction is changing. So, again, in case the speed is constant, $\vec a(t)$ is perpendicular to the curve. Its direction tells you which direction $\vec x(t)$ is turning, while its magnitude tells you how quickly $\vec x(t)$ is changing direction. This last quantity is a constant multiple of the curvature of $\vec x(t)$.
I think you might be confused because of the notation. If $\mathbb p$ is supposed to be the position, it should depend only on time, not on another position. That means $\mathbb p$ should be a function of one variable.
The differential equation is supposed to be $\frac{d\mathbb p}{dt} = \mathbb v(\mathbb p)$. The integral form is
$$
\mathbb p(t) = \mathbb p(0) + \int_{t_0}^{t} \mathbb v(\mathbb p(s)) ds,
$$
but this generally is not the method to find the solution.
To find $\mathbb p$ given $\mathbb p(0)$ and $\mathbb v$, it may help to break $\mathbb p(t)$ into $\mathbb p(t) = \langle p_x(t), p_y(t) \rangle$ and $\mathbb v(x, y)$ into $\mathbb v(x, y) = \langle v_x(x, y), v_y(x, y) \rangle$. The differential equation for the vector $\mathbb p$ can be written as a system of scalar differential equations:
\begin{align*}
\frac{dp_x}{dt} & = v_x(p_x, p_y) \\
\frac{dp_y}{dt} & = v_y(p_x, p_y).
\end{align*}
As an example, suppose $\mathbb v(x, y) = \langle 3x, xy \rangle$, i.e., $v_x(x, y) = 3x$ and $v_y(x, y) = xy$. Then you have the system
\begin{align*}
\frac{dp_x}{dt} & = 3p_x \\
\frac{dp_y}{dt} & = p_xp_y.
\end{align*}
The first equation can be solved independently for $p_x$, giving
$$p_x(t) = c_1e^{3t}$$
where $c_1$ is a constant. Substitute this into $p_x$ in the second equation to get
$$
\frac{dp_y}{dt} = c_1e^{3t}p_y.
$$
This equation is separable. The solution is
$$
p_y(t) = c_2e^{\frac{c_1}3 e^{3t}}
$$
where $c_2$ is another constant.
$c_1$ and $c_2$ can be determined once the initial condition is given.
If $\mathbb p(0) = \langle x_0, y_0\rangle$ is given, then
$$
\mathbb p(0) = \langle x_0, y_0\rangle =
\langle c_1, c_2e^{\frac{c_1}3} \rangle.
$$
It is easy to verify that $c_1 = x_0$ and $c_2 = y_0e^{-\frac{x_0}3}$. Therefore,
$$
\mathbb p(t) = \langle x_0e^{3t}, y_0e^{\frac{x_0}3\left(e^{3t} - 1\right)} \rangle
$$
Best Answer
The first step is already wrong. $\vec r(t)\cdot\vec r(t)$ is the squared length of $\vec r(t)$, which can vary arbitrarily with time.
And so from this incorrect premise, absurd things may be proven, like you just did.