checking that a map is a local isometry $$\varphi:\mathbb{R}^2\rightarrow \mathbb{R}^3:(x,y)\mapsto (\text{cos}(x),\text{sin}(x),y)$$
I wish to check that it is a local isometry, i.e.
$$ \varphi^*g_0 = g_0 \\(\varphi^*g_0)(\partial_x,\partial_y) = g_0(\varphi_*\partial_x,\varphi_*\partial_y) = g_0(-\text{sin}(x)\partial_{x^1}+\text{cos}(x)\partial_{x^2},\partial_{x^3}) = 0$$
My question is how do we get $$g_0(\varphi_*\partial_x,\varphi_*\partial_y) = g_0(-\text{sin}(x)\partial_{x^1}+\text{cos}(x)\partial_{x^2},\partial_{x^3})$$
Is the rule of calculation the following: $$\varphi_*\partial_x = \sum_i\partial_x\varphi^i\partial_{x^i}$$
 A: Consider the map $\varphi:\mathbf{R}^2\to\mathbf{R}^3$ given by $$(u,v)\mapsto(\cos(u),\sin(u),v).$$ I use the letters $(u,v)$ for the global coordinates of $\mathbf{R}^2$ to avoid confusion with the coordinates $(x,y,z)$ parameterizing $\mathbf{R}^3$.
Let $\partial_u=(1,0)$ and $\partial_v=(0,1)$ be the global frame of the tangent bundle, $T\mathbf{R}^2$. Note that since $\varphi$ is a mapping between submanifolds of Euclidean space that the differential in the standard coordinates is simply the Jacobian matrix. Now, at any point $(u,v)\in\mathbf{R}^2$,
$$\varphi_*(\partial_u)=d\varphi(\partial_u)=\frac{\partial\varphi}{\partial u}=(-\sin(u),\cos(u),0)=-\sin(u)\partial_x + \cos(u)\partial_y,$$
and
$$\varphi_*(\partial_v)=d\varphi(\partial_v)=\frac{\partial\varphi}{\partial v}=(0,0,1)=\partial_z.$$
Now the result immediately follows since by pulling back the metric of the cylinder back onto $\mathbf{R}^2$ via $\varphi$, we see that
$$(\varphi^*g)(\partial_u,\partial_v)=g(d\varphi(\partial_u),d\varphi(\partial_v))=g(-\sin(u)\partial_x + \cos(u)\partial_y,\partial_z)=0.$$
You can repeat similar computations to obtain $(\varphi^*g)(\partial_u,\partial_u)$ and $(\varphi^*g)(\partial_v,\partial_v)$.
