If $F$ is a field $a\in F$, if $(x-a)^n$ for $n\geq 2$ divides $r(x)$ then $r(a)=r^{\prime}(a)=0$ Statment:
If $F$ is a field  $a\in F$, if $(x-a)^n$  for $n\geq 2$ divides $r(x)$ then $r(a)=r^{\prime}(a)=0$.
Proof: Induction over $n$
Case $n=2$
Let $F$ be a field and $a\in F$ since $(x-a)^2 |r(x)$ then $(x-a)|r(x)$ and hence $r(a)=0$.
Now suppose that $r^{\prime}(a)\neq 0$ then it implies that $(x-a)$ isn´t a factor of $r^{\prime}(x)$ and hence by Euclid´s Algorithm we should have $$r^{\prime}(x)=(x-a)g(x)+u(x)$$ where $u=0$ or $deg(u)=0$.
If $u=0$ the proof finish and so let us assume that $deg(u)<1$ it is $u\in F$ and then from our relation $$r^{\prime}(x)-(x-a)g(x)=u$$
Now notice that in the RHS we have $u$ with degree $0$ and in the $RHS$ we have $deg(r^{\prime}(x)-(x-a)g(x))=0$ implies that $deg(r^{\prime}(x))=0$ or $deg((x-a)(g(x))=0$ since $r^{\prime}$ have degree $n-1$ and $(x-a)g(x)$ should have degree greater than $1$ it is impossible and
hence $r^{\prime}(x)=0$ and $$r(a)=r^{\prime}(a)=0$$ is true as well.
Now assume that the result is true for $n\leq k-1$ and we should prove the case $n=k$
Now consider $(x-a)^{k}|r(x)$ we must prove that $r(a)=r^{\prime}(a)=0$.
Since $(x-a)^k|r(x)$ then $r(a)=0$, consider $(x-a)^k=(x-a)^{k-1}(x-a)$ and since $(x-a)^k|r(x)$
then $(x-a)^{k-1}|r(x)$ and then by our hypothesis the equalty holds.
and the statment is true for $n\geq 2$.
I think that I made a mistake since the case $n=k$ was easy, where I made a mistake and how I should prove that.
 A: You should instead prove the standard properties of the derivative: if $p(x)=f(x)g(x)$, then
$$
p'(x)=f'(x)g(x)+f(x)g'(x)
$$
Then, by induction, if $q(x)=(x-a)^n$ and $n\ge2$, you have
$$
q'(x)=n(x-a)^{n-1}
$$
Since $r(x)=(x-a)^ns(x)$, you have
$$
r'(x)=n(x-a)^{n-1}s(x)+(x-a)^ns'(x)
$$
and direct substitution shows that $r'(a)=0$.
This also allows to prove the converse, namely that if $r(a)=r'(a)=0$, then $r(x)$ is divisible by $(x-a)^2$ and hence by $(x-a)^n$ for some maximal $n\ge2$.
Indeed, you can write $r(x)=(x-a)^2s(x)+v(x)$ and
$$
r'(x)=2(x-a)s(x)+(x-a)^2s'(x)+v'(x)
$$
Thus $v(a)=0$ and $v'(a)=0$. Since the degree of $v$ is at most $2$, we get that $v(x)=0$.

How to prove it? By properly defining the derivative. Let $f(x)\in F[x]$. Then add an indeterminate $h$ and define
$$
f_0(x,h)=f(x+h)-f(x)\in F[x,h]
$$
Since $f_0(x,0)=0$, we know that $f_0(x,h)$ is divisible by $h$ and so we can write
$$
f_0(x,h)=hf_1(x,h)
$$
Now define $f'(x)=f_1(x,0)$. Proving the standard properties and deriving that if $f(x)=a_0+a_1x+\dots+a_kx^k$ then
$$
f'(x)=a_1+a_2x+\dots+ka_kx^{k-1}
$$
is straightforward.
