If we have a function $f$ such that $f:R\to R^+$ where $f(x) = 3^{-x}$, it's graph will be like that
And we have another function $g$ such that $g: R\to R^+$ where $g(x) = f(x+1) = 3^{-x+1} = 3^{1-x}$
And because $g(x) = f(x+1)$ so the curve of $g$ is an image of the curve of $f$ by translation 1 unit to the left
But it turns out that the translation is 1 unit to the right and not to the left
The graph of $f(x+1)$ is supposed to be the same as that of $f(x)$ by translation towards the negative part of x-axis.Why isn't that the case here?
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then $$ g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}. $$