1. Given real numbers $x, y$ satisfy the equation
$$
x^{2}-3 x y+3 y^{2}+4 x-18 y+52=0 \text {. }
$$

then the units digit of $y^{x}$ is $\qquad$ .
$=1.4$.
From the given equation, we have
$$
x^{2}-(3 y-4) x+\left(3 y^{2}-18 y+52\right)=0 \text {. }
$$

Since $x$ is a real number, then
$$
\Delta=(3 y-4)^{2}-4\left(3 y^{2}-18 y+52\right) \geqslant 0 \text {, }
$$

which simplifies to $-3(y-8)^{2} \geqslant 0$.
Thus, $y=8$.
Substituting $y=8$ into the original equation, we get $x^{2}-20 x+100=0$.
Solving this, we find $x=10$.
Therefore, $y^{z}=8^{10}=2^{30}=\left(2^{4}\right)^{7} \times 2^{2}=4 \times 16^{7}$.
Since the units digit of $16^{7}$ is 6, the units digit of $4 \times 16^{7}$ is 4.