I'm told that the Legendre equation
$$(1-x^2)\frac{d^2P_n}{dx^2}-2x\frac{dP_n}{dx} +n(n+1)P_n=0$$
can be transformed by a change of variables $\tilde{x}=\frac{1-x}{1+x}$ into the hypergeometric equation. So to try to perform that change of variables I write
$$\tilde{x}+x\tilde{x}=1-x \qquad \Rightarrow \qquad x(\tilde{x}+1)=1-\tilde{x} \qquad \Rightarrow $$ $$x=\frac{1-\tilde{x}}{1+\tilde{x}}$$
Next I want to know the relationship between derivatives, so
$$\frac{dx}{dx} = \frac{-\frac{d\tilde{x}}{dx}(1+\tilde{x})-(1-\tilde{x})\frac{d\tilde{x}}{dx}}{(1+\tilde{x})^2} \qquad \Rightarrow $$ $$1=\frac{-2\frac{d\tilde{x}}{dx}}{(1+\tilde x)^2}$$
So $\frac{d\tilde x}{dx} = -(1+\tilde x)^2/2$.
So now we can write $P_n(x) = y(\tilde x)$ and
$$\frac{dP_n}{dx} = \frac{d}{dx}y(\tilde x) = y'(\tilde x)\cdot -(1+\tilde x)^2/2$$
$$\frac{d^2P_n}{dx^2} = -\frac{1}{2}\left( 2(1+\tilde x)\frac{d\tilde x}{dx}y'(\tilde x) + (1+\tilde x)^2y''\frac{d\tilde x}{dx} \right)$$ $$=\frac{(1+\tilde x)^3}{4}\left(2y'+(1+\tilde x)y''\right)$$
After substituting these into the Legendre equation, though, doesn't seem to transform it into anything nice or useful. It's making me wonder if I've done something wrong.
The problem here is that you have to set $$ P_n(x) = \frac{y(\tilde{x})}{(1-\tilde{x})^n}. $$ Do this and it should all work out.
Alternatively you could set $\tilde{x} = (1-x)/2$, which also gives the hypergeometric equation and is algebraically simpler. The form for $\tilde{x} = (x-1)/(x+1)$ then follows from the Pfaff transformation.