Pregunta: 2Find delzdelx whenxe2y+yz+ze3x=0.delzdelx=e3x-ye2y+3ze3xdelzdelx=e2y+3ze3xe3x+ydelzdelx=-e3x+ye2y+3ze3xdelzdelx=-e2y+3ze3xe3x+ydelzdelx=-e2y+3ze3xe3x-ydelzdelx=e3x+ye2y+3ze3x

$2$Find $\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{x}}$ when

$x{e}^{2y}+yz+z{e}^{3x}=0.$

$\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{x}}=\frac{e^{3x}-y}{e^{2y}+3z{e}^{3x}}$

$\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{x}}=\frac{e^{2y}+3z{e}^{3x}}{e^{3x}+y}$

$\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{x}}=-\frac{e^{3x}+y}{e^{2y}+3z{e}^{3x}}$

$\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{x}}=-\frac{e^{2y}+3z{e}^{3x}}{e^{3x}+y}$

$\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{x}}=-\frac{e^{2y}+3z{e}^{3x}}{e^{3x}-y}$

$\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{x}}=\frac{e^{3x}+y}{e^{2y}+3z{e}^{3x}}$