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.

A conic curve, is the intersection of the 3D cone $p_1z^2=p_2x^2+p_3y^2$ with the 3D plane $p_4x+p_5y+p_6=0$, which takes the general form as stated in the OP. Now, a point can also be regarded as a conic curve (or conic, simply) in a sense that it is the intersection of the 3D plane with the vertex of the 3D cone, i.e. the point in which, the two halves of the cone meet. An example is the intersection of $z^2=x^2+y^2$ with $2z=x+y$. Hence, your approach is correct. Another way for justifying the result, is to note that any single point is an ellipse with zero diameters and ellipse is a conic curve.

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.