$$f(x,y)=3x^2+5y^2-2x+3y+7=0\\f'x=6x-2=0\\x=\frac {1}{3}\\f'y=10y+3=0\\ y=\frac {-3}{10} \\ O(\frac {1}{3},\frac {-3}{10})$$
As you see, we've used the partial derivative to find the center point
of an ellipse.We did partial derivative with respect to x and did the
same with respect to y.
I know that if we have a function with two variables like z=f(x,y) (in
three dimensional space) it gives us a surface.Then we can think of
the partial derivative as a tangent plane to that specific point.But
here(ellipse equation) we've got a curve equation on two dimensional
space and also it's not a function(!) it's an equation(!) how are we
allowed to take partial derivative of it, how is it possible to find the center of an ellipse by this method and WHAT IS THE GEOMETRIC
POINT OF VIEW ( like tangent plane in 3 dimensional space) of getting
these partial derivatives in two dimensional space.
Thanks
$ f (x, y) $ is a function of $ x $ and $ y $ whatever the value of $ f $ is, so I don't see why there is any issue in taking the partial derivatives of $ f $. It just so happens that when $ f=0$ the problem collapses to two dimensions.
We can find the center if the ellipse using the partial derivatives of the ellipse since the equation of an ellipse is
$\frac {(x-h )^2}{a} + \frac {(y-k )^2}{b} = 1$
If we take the partial derivative of the LHS wrt to $ x $ or $ y $ (which $=0$ since we are differentiating a constant), you can see how solving the equation that we get yields the coordinates of the center.
As for geometric intuition, well, $ f=0$ is simply the $ xy $ -plane. The partial derivatives of $ f $ wrt $ x $ or $ y $ at a point $(a, b) $ are tangents to the curves lying in the planes $ x=a $ and $ y=b $ but these are all flat anyway so the tangents are all zero