Answer
dz=0.12
Work Step by Step
We are given $z=f(x,y)=2x^{2}+4xy+y^{2}$
$f_{x}(x,y)=4x+4y$
$f_{y}(x,y)=4x+2y$
By the definition $dz=f_{x}(x,y).dx+f_{y}(x,y)dy=$
$dz=(4x+4y)dx+(4x+2y)dy$
$dz=[4\times5+4\times(-1)]\times0.03+[4\times5+2\times(-1)]\times(-0.02)=0.12$