The Euclidean metric after the mapping $f:\Bbb R^2 \to \Bbb R^{2+}$ with $f(x,y)=(e^x,e^y),$ is what? I started with $\Bbb R^2$ and the Euclidean metric $ds^2=dx^2+dy^2.$ Then used a map $f:\Bbb R^2\to\Bbb R^{2+}$ with $f(x,y)=(e^x,e^y).$

How do you write down the new metric after the map?

I know that under the map, the origin $(0,0)\mapsto(1,1).$ So $(1,1)$ is the new origin, in the image of the map.
I searched in "An Introduction to Manifolds" by Tu, to try to figure out how to write down the new metric after the map. I searched in the section about "push-forwards" because I think that's what I need.
So from following my text supplemented with a wikipedia article, if the tangent space is defined with differentiable curves $g_n(x)=\pm\frac{n}{x}$ and $df_p:T\Bbb R^2\to T\Bbb R^{2+}$ then $df_p(g'_n(0))=(f \circ g_n)'(0).$ I'm stuck at this part.
 A: Your question "How do you write down the new metric after the map?" is difficult to understand. It seems that you want to transport the Euclidean metric $ds^2=dx^2+dy^2$ into the first quadrant of the $(u,v)$-plane, such that for all curves $\gamma$ in the $(x,y)$-plane we have $L_{uv}\bigl(f(\gamma)\bigr)=L_{xy}(\gamma)$.
Any $ds^2$ in the $(u,v)$-plane will be of the form
$$ds^2=E(u,v)du^2+2F(u,v)dudv+G(u,v)dv^2$$
with certain functions $E$, $F$, $G$. (This is the oldfashioned way to write the $g_{ik}$ tensor.) The length of a curve $$\alpha:\quad t\mapsto\bigl(u(t),v(t)\bigr)\qquad(a\leq t\leq b)$$ then computes to
$$L_{uv}(\alpha)=\int_a^b\sqrt{E(\alpha(t))u'^2(t)+2F(\alpha(t))u'(t)v'(t)+G(\alpha(t))v'^2(t)}\>dt\ .\tag{1}$$
Now we want
$$L_{uv}(\alpha)=L_{xy}\bigl(f^{-1}(\alpha)\bigr)\ .$$
As $$f^{-1}(\alpha):\quad t\mapsto \bigl(x(t),y(t)\bigr):=\bigl(\log u(t),\log v(t)\bigr)$$
this means that
$$L_{uv}(\alpha)=\int_a^b\sqrt{x'^2(t)+y'^2(t)}\>dt=\int_a^b\sqrt{{u'^2(t)\over u^2(t)}+{v'^2(t)\over v^2(t)}}\>dt\ ,\tag{2}$$
for any curve $\alpha$.
Comparing $(2)$ with $(1)$ we see that for the given map $f$ we need
$$E(u,v)={1\over u^2},\quad F(u,v)\equiv0,\quad G(u,v)={1\over v^2}\ ,$$
so that the $ds^2$ in the $(u,v)$-plane will be
$$ds^2={1\over u^2}du^2+{1\over v^2}dv^2\ .$$
