What does partial differentiation give for a second degree equation which doesn't represent a conic? Question says:
Find real solution to the equation $$3x^{2}+3y^{2}-4xy+10x-10y+10=0.$$
My first thought was to treat it as a general conic ( $ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $) but when I do so, its discriminant is lesser than $0$. And on plotting on geogebra, also it only shows a point.
And this is the same point we get after doing partial differentiation and solving for $x,y$. Generally for a conic, this $(x,y)$ represents the center of the conic. But for equations like these which don't represent a conic, how does doing partial differentiation and then solving, gives us the integer solution to it. If it is center, then there should be a conic too.
My friends told that conic is in a complex plane, but shouldn't center should also be in complex plane? Sorry Idk much about conics which exist in both planes, so maybe what I wrote in last 2 lines is completely wrong. Pardon me for that.

Multiplying by $\,3\,$ then "completing the square" for the quadratic in $\,x\,$:
$$
\begin{align}
& 9x^2+9y^2-12xy+30x-30y+30
\\ = \;\; &9x^2 - 6\, (2y-5)\,x \color{red}{+(2y-5)^2-(2y-5)^2} +9y^2-30y+30
\\ = \;\;&\big(3x-(2y-5)\big)^2 + 5y^2-10y+5
\\ = \;\;&(3 x - 2 y + 5)^2 + 5 (y-1)^2
\end{align}
$$
It follows that the only real solution is $\,y=1, x=-1\,$.

[ EDIT ] $\;$ The final equation can be written as $\,x'^{\,2}+ y'^{\,2}=0\,$ with $\,x' = 3x-2y+5\,$ and $\,y' = \sqrt{5} y-\sqrt{5}\,$, which is a degenerate conic, specifically two intersecting complex lines $\,x' \pm i\, y' = 0\,$ with the single common real point at $\,x' = y' = 0\,$. This is consistent with OP's findings of discriminant $\,\le 0\,$ and center of the conic at the unique real point.
