Why does this formula yield a clockwise rotation for curves defined by a cartesian equation, and a counterclockwise rotation in parametric form? Here is the reasoning I've attemped in order to espablish a formula for the counterclockwise rotation of a curve.
Let $M=(x,y)$ be any point on the plane, associated with a cartesian coordinate system $<0, \vec { i}, \vec{j}>$.
The position vector of ths point is $\vec {OM} = \vec A+ \vec B= x\vec { i} + y \vec{j} $
Rotating point $M$ counterclockwise by a positve angle $\alpha$ amounts to rotate counterclockwise  the components of its postion vector by the same angle.
So , with $M'$ the image of $M$ under ths rotation, we will have :
$\vec {OM'} = \vec {A'}+\vec {B'}$ with  $\vec {A'}$, the image of $\vec {A}$ and $\vec {B'}$ , the image of vector $\vec {B}$.
Now ,
$\vec {A'}= x cos (\alpha) \vec i + x sin (\alpha) \vec j$.
$\vec {B'}= y cos (\alpha + \pi /2) \vec i + x sin (\alpha+ \pi/2) \vec j $
$= y ( - sin (\alpha) )\vec i + x cos(\alpha) \vec j$
$= -y  sin (\alpha) )\vec i + x cos(\alpha)\vec j$.
Therefore ,
$\vec{OM'} =  \vec {A'} + \vec {B'} = \Large{[} x cos (\alpha) - y  sin (\alpha)\Large{]} \vec i + \Large{[}x sin( \alpha) + y cos(\alpha)\Large{]} \vec j$,
implying that
$M' = (x cos (\alpha) - y  sin (\alpha) , x sin( \alpha) + y cos(\alpha))$.
My problem is that :
(1) when I apply this reasoning to a parametric curve, it seems to work , that is, if I have a curve $C= < f(t), g(t)> $ and ask for
$$C' = <f(t) cos (\alpha) - g(t)  sin (\alpha) , f(t) sin( \alpha) + g(t) cos(\alpha)>$$
what I get is that  $C'$ rotates counterclockwise, as desired.
Example with $\alpha = 2 $ rd: https://www.desmos.com/calculator/daaouk87yw
(2) but if I substitute the coordinates of a generic $M'$ point in a cartesian " $x-y$" equation, the curve rotates clockwise.
Example : https://www.desmos.com/calculator/vm5pbtxzge
I'm sure some confusion prevents me from understanding what is happening, but I can't manage to see which one.
 A: Okay I think if you write things in the notation of matrices it becomes a little easier to see what's happening. So we're going to use only the position vectors $\begin{bmatrix}x \\ y \end{bmatrix}$ to represent the terminal point (x,y). And then we can think of counterclockwise rotation around the origin by angle $\alpha$ as applying the rotation matrix
$R=\begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha &  \cos \alpha \end{bmatrix}$
So $\begin{bmatrix}x \\ y \end{bmatrix}$ rotated by $\alpha$ is
$\begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha &  \cos \alpha \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}=
\begin{bmatrix}
x\cos \alpha  -y\sin \alpha  \\
x\sin \alpha  +y\cos \alpha
\end{bmatrix}$
which agrees with the formulas you gave.
Now lets consider rotating a set of vectors defined parameterically
$\left\{\begin{bmatrix}x \\ y \end{bmatrix} \mid x=f(t), y=g(t)\right\}$ by $\alpha$. The resulting set will be just multiplying each set by $R$, that is $\left\{R\begin{bmatrix}x \\ y \end{bmatrix}\mid x=f(t), y=g(t)\right\}$
If we do the same for a set of vectors defined implicitly
$\left\{\begin{bmatrix}x \\ y \end{bmatrix} \mid f(x,y)=0\right\}$
multiplying by $R$ still works
$\left\{R\begin{bmatrix}x \\ y \end{bmatrix} \mid f(x,y)=0\right\}$
However you are looking for a new function $F(x,y)$ such that the set of rotated points is
$\left\{\begin{bmatrix}x \\ y \end{bmatrix} \mid F(x,y)=0\right\}$
Edit:
Okay to see why this is not the same thing it may be worth referring to these new rotated variables by a different notation. Let's use capital letters to define the rotated coordinates:
$\begin{bmatrix}X \\ Y \end{bmatrix}=R \begin{bmatrix}x \\ y \end{bmatrix}=\begin{bmatrix}
x\cos \alpha  -y\sin \alpha  \\
x\sin \alpha  +y\cos \alpha
\end{bmatrix}$
So to summarize you start with a set defined:
$\left\{\begin{bmatrix}x \\ y \end{bmatrix} \mid f(x,y)=0\right\}$ where you could take $f(x,y)=\frac{y^2}{5^2}+\frac{x^2}{3^2}-1$ to get an ellipse like your original examples.
and you want to convert to a set defined by:
$\left\{\begin{bmatrix}X \\ Y \end{bmatrix} \mid F(X,Y)=0\right\}$.
It would be nice to use the original function $f(x,y)$ to define the relationship but it only works for the unrotated vectors. So that means you need a way to go from the rotated vectors $\begin{bmatrix}X \\ Y \end{bmatrix}$ to the unrotated vectors $\begin{bmatrix}x \\ y\end{bmatrix}$ in order to put in your $f(x,y)$ to see if they match they satisfy $f(x,y)=0$.
Since $\begin{bmatrix}X \\ Y \end{bmatrix}=R \begin{bmatrix}x \\ y \end{bmatrix}$, you can apply the inverse to get $\begin{bmatrix}x \\ y \end{bmatrix}=R^{-1}\begin{bmatrix}X \\ Y \end{bmatrix}$. Naturally the inverse of a rotation is simply rotating by the same amount in the opposite direction. So the expression you would plug into the original defining equation would be rotated in the opposite direction of the rotation of the points.
A: We'll compare the two operations you describe. Instead of reasoning with coordinates, let's use directly the point $M$, and the operation $R$ applied to $M$. $R$ can be a rotation, or any other bijective function.

*

*In the first case, you directly change the parametric equation $M = F(t)$ that expresses $M$ according to some parameter $t$. This becomes $M' = R \circ F (t)$.

*In the second case, you transform the usage of $M$ coordinates in an equation they verify. If the equation is $E(M) = 0$, it becomes $E(R(M''))=0$.

$R$ being bijective, $E(R(M''))=0$ is equivalent to $M=R(M'')$.
While in the first case $M'=R(M)$.
We can see the role of $R$ is reversed; hence what you remarked: if you express $M''=S(M)$, we see that $S=R^{-1}$.
This is very intuitive, in fact. Let's say it with a scaling operation; compare:

*

*"I'll give you twice the amount $M$".

*"The amount I'll give you is such that, when multiplied by two, it is the same as $M$".

Another similar situation happens when transforming a vector, versus transforming the vector base: that's covariance and contravariance.
