suppose $\epsilon > 0$ is given and choose $\delta = \min \{1, \epsilon/21 \}.$ now if $0 < |x-1|, |y-1|, |z-1| < \delta,$ then $0 because $\delta \leq 1.$
we also have: $yx^2+2xz^2 - 3 = (y-1)x^2 + (x-1)(x+1+2z^2) + 2(z-1)(z+1).$ thus: $|yx^2+2xz^2-3| < 4 \delta + 11 \delta + 6 \delta = 21 \delta < \epsilon.$