Answer
$$\frac{\partial z}{\partial s}=\frac{3\sin t-2t\sin s}{3s\sin t+2t\cos s}$$
$$\frac{\partial z}{\partial t}=\frac{3s\cos t+2\cos s}{3s\sin t+2t\cos s}$$
Work Step by Step
The partial derivative with respect to $s$ is:
$$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}=
\frac{\partial }{\partial x}(\ln(3x+2y))\frac{\partial}{\partial s}(s\sin t)+\frac{\partial}{\partial y}(\ln(3x+2y))\frac{\partial}{\partial s}(t\cos s)=
\frac{1}{3x+2y}\frac{\partial}{\partial x}(3x+2y)\cdot \sin t+\frac{1}{3x+2y}\frac{\partial}{\partial y}(3x+2y)\cdot t\cdot(-\sin s)=
\frac{3\sin t-2t\sin s}{3x+2y}$$
Now we will express solution in terms of $s$ and $t$:
$$\frac{\partial z}{\partial s}=\frac{3\sin t-2t\sin s}{3x+2y}=
\frac{3\sin t-2t\sin s}{3s\sin t+2t\cos s}$$
The partial derivative with respect to $t$ is:
$$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}=
\frac{\partial }{\partial x}(\ln(3x+2y))\frac{\partial}{\partial t}(s\sin t)+\frac{\partial}{\partial y}(\ln(3x+2y))\frac{\partial}{\partial t}(t\cos s)=
\frac{1}{3x+2y}\frac{\partial}{\partial x}(3x+2y)\cdot s\cos t+\frac{1}{3x+2y}\frac{\partial}{\partial y}(3x+2y)\cdot\cos s=
\frac{3s\cos t+2\cos s}{3x+2y}$$
Expressing solution in terms of $s$ and $t$ we get:
$$\frac{\partial z}{\partial t}=\frac{3s\cos t+2\cos s}{3x+2y}=
\frac{3s\cos t+2\cos s}{3s\sin t+2t\cos s}$$
