Solución
∂
z
∂
r
=
3
e
3
,
∂
z
∂
θ
=
(
2
−
4
3
)
e
3
\frac{\partial z}{\partial r} = \sqrt{3}e^{\sqrt{3}},\; \frac{\partial z}{\partial \theta} = (2-4\sqrt{3})e^{\sqrt{3}}
∂
r
∂
z
=
3
e
3
,
∂
θ
∂
z
=
(
2
−
4
3
)
e
3