Consider $(S^1, g)$ where $S^1$ is the unit circle and g is a metric. Now consider the metric $$ \tilde g := f g $$ where f is a smooth positive function. Since in 1 dimension this is the only smooth deformation that is possible it seems to me that any deformation would necessarily be conformal.
Since I am asked to show a certain equality holds for deformations of a metric on $S^1$ that are not necessarily conformal I wonder where my reasoning is going wrong. Does the task make sense ?
On
Given the natural global coordinate vector field $\partial\theta$, Riemannian metrics on $S^1$ can be identified with positive functions. In particular, as you observe, the ratio of two positive functions is another positive function, so the only smooth deformations are conformal.
Another way of saying this is that the only angles between vectors on $S^1$ are $0$ and $\pi$. This is a discrete set, so any smooth deformation of the metric necessarily preserves angles between vectors.
As you say any two metrics on $\mathbb{S}^1$ are conformal. Let $V$ a vector field tangent to $\mathbb{S}^1$ and nowhere zero. Then, if $g$ and $\tilde{g}$ are two Riemannian metrics it must be $\tilde{g}(V,V)>0$ and $g(V,V)>0.$ If we define $f=\dfrac{\tilde{g}(V,V)}{g(V,V)}$ one gets $\tilde{g}=fg$ with $f>0.$ That is, $g$ and $\tilde{g}$ are conformal.