We have

$x=r \sin θ \cos⁡ ∅, \\ y=r \sin θ \sin⁡ ∅ \\ z=r \cos ⁡θ$

∴According to Formula

$J=\dfrac {d(x,y,z)}{d(r,\theta,\phi)} =\begin{vmatrix} \dfrac {dx}{dr} & \dfrac {dx}{d\theta} & \dfrac {dx}{d\phi} \\ \dfrac {dy}{dr} & \dfrac {dy}{d\theta} & \dfrac {dy}{d\phi} \\ \dfrac {dz}{dr} & \dfrac {dz}{d\theta} & \dfrac {dz}{d\phi}\\ \end {vmatrix} $

Differentiating x w.r.t $r,θ$ & $∅$ We get $\dfrac {∂x}{∂r} =\sin θ \cos ⁡∅ \ \dfrac {∂x}{∂θ} = r \cos θ \cos ⁡∅ \ \dfrac {∂x}{∂∅}= r\sin θ(-\sin ∅)$ Similarly $$\dfrac {∂y}{∂r} =\sin θ \sin ⁡∅ \ \dfrac {∂y}{∂θ} = r \cos θ \sin ⁡∅ \ \dfrac {∂y}{∂∅} = r \sin θ \cos ⁡∅ \ \dfrac {∂z}{∂r} = \cos ⁡θ \ \dfrac {∂z}{∂θ} = -r \sin ⁡θ \ \dfrac {∂z}{∂∅}=0 $$

$$∴J= \begin{vmatrix} \sin θ \cos ⁡∅ & r \cos θ \cos ⁡∅ & -r \sin θ \sin ⁡∅ \\ \sin⁡ θ \sin⁡ ∅ & r \cos θ \sin ⁡∅ & r \sin θ \cos ⁡∅ \\ \cos\theta & -r\sin \theta & 0\\ \end {vmatrix} $$

$=\sin⁡ θ \cos⁡ ∅ [0-(-r^2 \sin^2⁡ θ \cos ⁡∅ )]-r \cos⁡ θ \cos ∅[0-r \cos θ \sin⁡ θ \cos ⁡∅] - r \sin⁡ θ \sin ∅[-r \sin^2⁡ θ \sin ∅- r\cos^2 θ \sin ⁡∅ ] \\ =r^2 \sin^3⁡ θ \cos^2⁡ ∅+ r^2 \sin⁡ θ \cos^2 θ \cos^2⁡ ∅+ r^2 \sin^3⁡ θ \sin^2⁡ ∅+ r^2 \sin θ \cos^2 \sin^2 ∅ \\ =r^2 \sin θ \cos^2 ∅[\sin^2 θ + \cos^2 ⁡θ ] + r^2 \sin θ \sin^2 ∅[\sin^2 θ + \cos^2 ⁡θ ] \\ =r^2 \sin θ\cos^2 ∅ × 1+ r^2 \sin⁡ θ \sin^2⁡ ∅ × 1 \\ = r^2 \sin ⁡θ[\cos^2⁡ ∅ + \sin^2 ⁡∅ ] \\ =r^2 \sin θ×1 \\ J=r^2 \sin ⁡θ\\ ∴J^1= \dfrac 1J = \dfrac 1{r^2 \sinθ} \\ ∵J.J^1=1 $

Hence $\dfrac {∂(r,θ,∅)}{∂(x,y,z)} = \dfrac 1{r^2 \sin θ }$