Are you smarter than me? 53

$3x^{2} - 6x + 3 + 3y^{2} - 6y + 3 + 4 +4x - 4y + 4xy = 0$

$= 3((x - 1)^{2}) + 3(( y - 1)^{2}) + 4(x - 1)(y - 1) = 0$

$=(x - 1)^{2} + (y - 1)^{2} +2[(x - 1)^{2} + (y - 1)^{2} + 2(x - 1)(y - 1)] = 0$

$( x - 1)^{2} + (y - 1)^{2} + 2[(x - 1) + (y - 1)]^{2} = 0$

$(x , y ) =( 1 , 1 )$

Please upvote my solution

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Let $x+y=a$ and $xy=b$

Then we'll get $3((x+y)^2-2xy)+4xy-10(x+y)+10=0$

$\Rightarrow 3a^2-6b+4b-10a+10=0$

$\Rightarrow (a)(3a-10)=(2)(3b-5)..........(1)$

Now as it has integral solution therefore $a=2$ (By comparing both sides)

Similarly substituting $a=2$ in $.........(1)$ we get $b=1$

hence $x+y=\huge\boxed{2}$

We can solve further and get $x=y=1$