Since there are two constraints, we must use $$\nabla f(x,y,z)=\lambda\nabla g + \mu \nabla h.$$
We have
$f(x,y,z)=xy+z^2$
$g(x,y,z)=2x-y$
$h(x,y,z)=x+z$
So we get
\begin{align}
f_x=\lambda g_x+\mu h_x &\implies y=2\lambda + \mu \tag{1} \\
f_y=\lambda g_y+\mu h_y &\implies x=-\lambda \tag{2} \\
f_z=\lambda g_z+\mu h_z &\implies 2z=\mu \tag{3}
\end{align}
with your constraint equations
\begin{align}
2x-y&=0 \tag{4} \\
x+z &=0 \tag{5}
\end{align}
Utilizing the system of equations $(1)$ to $(5)$, can you take it from here?
Since there are two constraints, we must use $$\nabla f(x,y,z)=\lambda\nabla g + \mu \nabla h.$$ We have
So we get \begin{align} f_x=\lambda g_x+\mu h_x &\implies y=2\lambda + \mu \tag{1} \\ f_y=\lambda g_y+\mu h_y &\implies x=-\lambda \tag{2} \\ f_z=\lambda g_z+\mu h_z &\implies 2z=\mu \tag{3} \end{align} with your constraint equations \begin{align} 2x-y&=0 \tag{4} \\ x+z &=0 \tag{5} \end{align} Utilizing the system of equations $(1)$ to $(5)$, can you take it from here?