To OP: You may have got your answer, I am adding mine solely to suggest another way to think.

You are smoothly mapping (-1,1) to (0,\infty), to make matters simple let’s think increasing functions only. So f(x)—>0 as x—>-1, and f(x)—>\infty as x—>1.

Now you needn’t worry about f(0)=1 constraint, because in case our function has f(0)=c, we can just replace g(x)=f(x)/c that would satisfy all our requirements of correct limits. So let’s concentrate on the rest.

One readymade function which increases from zero to infinity in some range is tan(x), which takes (0,π/2) to (0,\infty). So if you can take the interval (-1,1) to (0,π/2) by a function h; then you are done; a candidate for f would be tan(h(x)).

Finding a smooth h is almost trivial, since you can fit a line h(x)=ax+b. With the conditions h(-1)=0 and h(1)=π/2, you can solve for a,b; solutions: >!(a=π/4, b=π/4)!<

So our f(x) is tan(ax+b), and c=f(0)=tan(b). So final answer would be g(x)=tan(ax+b)/tan(b).

Happy calculusing!