Find real solution to the equation3x2+3y2−4xy+10x−10y+10=0.What does partial differentiation give for a second degree equation...

Step 1
A conic curve, is the intersection of the 3D cone
${\displaystyle p_1{z}^2=p_2{x}^2+p_3{y}^2}$
with the 3D plane
${\displaystyle p_{4}x+p_{5}y+p_6=0}$
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
${\displaystyle z^2=x^2+y^2}$ with ${\displaystyle 2z=x+y}$
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.

Step 1
Multiplying by 3 then "completing the square" for the quadratic in x:
${\displaystyle 9x^2+9y^2-12xy+30x-30y+30}$
${\displaystyle =9x^2-6(2y-5)x+{(2y-5)}^2-{(2y-5)}^2+9y^2-30y+30}$
${\displaystyle ={(3x-(2y-5))}^2+5y^2-10y+5}$
${\displaystyle ={(3x-2y+5)}^2+5{(y-1)}^2}$
It follows that the only real solution is ${\displaystyle y=1,x=-1}$
Step 2
The final equation can be written as
${\displaystyle x^{\prime 2}+y^{\prime 2}=0}$ with ${\displaystyle x^{\prime }=3x-2y+5}$ and ${\displaystyle y^{\prime }=\sqrt{5}y-\sqrt{5}}$,
which is a degenerate conic, specifically two intersecting complex lines ${\displaystyle x^{\prime }\pm i{y}^{\prime }=0}$ with the single common real point at ${\displaystyle x^{\prime }=y^{\prime }=0}$. This is consistent with OP's findings of discriminant ${\displaystyle \le 0}$ and center of the conic at the unique real point.