Deriving the round metric I want to derive the round metric $g=d\theta^{\,2}+\sin\left(\theta\right)^2d\phi^{\,2}$ but I cannot get the correct answer. I know that the metric in cartesian coordinates is $g=dx^2+dy^2$. I've used the formula $dx=\frac{dx}{d\theta}d\theta+\frac{dx}{d\phi}d\phi$ and $dy=\frac{dy}{d\theta}d\theta+\frac{dy}{d\phi}d\phi$. From
$x=\sin\left(\theta\right)\cos\left(\phi\right)$ and $y=\sin\left(\theta\right)\sin\left(\phi\right)$,
I find that
$\frac{dx}{d\theta}=\cos\left(\theta\right)\cos\left(\phi\right)$,
$\frac{dx}{d\phi}=-\sin\left(\theta\right)\sin\left(\phi\right)$,
$\frac{dy}{d\theta}=\cos\left(\theta\right)\sin\left(\phi\right)$,
$\frac{dy}{d\phi}=\sin\left(\theta\right)\cos\left(\phi\right)$.
This leads to $g=dx^2+dy^2=\cos\left(\theta\right)^2d\theta^{\,2}+\sin\left(\theta\right)^2d\phi^{\,2}$ which is not correct.
edit: metric was typed incorrectly
 A: calculate 
$$g_{\theta\theta}=\frac {\partial x}{\partial \theta }\frac {\partial x}{\partial \theta }+\frac {\partial y}{\partial \theta }\frac {\partial y}{\partial \theta }+\frac {\partial z}{\partial \theta }\frac {\partial z}{\partial \theta }$$
$$g_{\theta\phi}=\frac {\partial y}{\partial \theta }\frac {\partial y}{\partial \phi}+\frac {\partial x}{\partial \theta }\frac {\partial x}{\partial \phi }+\frac {\partial z}{\partial \theta }\frac {\partial z}{\partial \phi }$$
$$g_{\phi\theta}=\frac {\partial x}{\partial \phi}\frac {\partial x}{\partial \theta }+\frac {\partial y}{\partial \phi }\frac {\partial y}{\partial \theta }+\frac {\partial z}{\partial \phi }\frac {\partial z}{\partial \theta }$$
$$g_{\phi\phi}=\frac {\partial y}{\partial \phi }\frac {\partial y}{\partial \phi }+\frac {\partial x}{\partial \phi }\frac {\partial x}{\partial \phi}+\frac {\partial z}{\partial \phi }\frac {\partial z}{\partial \phi }$$
now plug it in the matrix $$g=\begin{pmatrix}g_{\theta \theta }&&g_{\theta \phi}\\g_{ \phi \theta} &&g_{\phi\phi}\end{pmatrix}$$
A: Let $\Psi(\theta, \phi): (\sin\left(\theta\right)\cos\left(\phi\right), \sin\left(\theta\right)\sin\left(\phi\right), \cos(\theta) )$. Then the coefficients of the metric are given by (The dot denotes the dot product, and using the Pythagorean identity $\sin^2\alpha + \cos^2\alpha=1$):
$\frac{\partial\Psi}{\partial \theta}\cdot\frac{\partial\Psi}{\partial\theta}=\left(\cos\left(\theta\right)\cos\left(\phi\right), \cos\left(\theta\right)\sin\left(\phi\right), -\sin\theta\right)\cdot \left(\cos\left(\theta\right)\cos\left(\phi\right), \cos\left(\theta\right)\sin\left(\phi\right), -\sin\theta\right)= 1$,
$\frac{\partial\Psi}{\partial \theta}\cdot\frac{\partial\Psi}{\partial \phi}=\left(\cos\left(\theta\right)\cos\left(\phi\right), \cos\left(\theta\right)\sin\left(\phi\right), -\sin\theta\right)\cdot \left(-\sin\theta\sin\phi,\sin\theta\cos\phi,0\right)=0$,
$\frac{\partial\Psi}{\partial\phi}\cdot\frac{\partial\Psi}{\partial\phi}= \left(-\sin\theta\sin\phi,\sin\theta\cos\phi,0\right)\cdot \left(-\sin\theta\sin\phi,\sin\theta\cos\phi,0\right)=\sin^2\theta$
