Vrftjkry

Suppose a function
"f (xy)", then does it equate to "f ' (xy) . x" when partial differentiating wrt y, and f '(xy).y when partial differentiating wrt x?

Cause I don't get it why differentiating f with x and y would both yield f ', any help?

Based on the comments, my guess is that the Newton notation is getting in the way a little. It might be more clear what is going on if we are a little more explicit with the functional composition, and if we use the Leibniz notation instead.

Suppose that
$$ f : mathbb{R} to mathbb{R}$$
is a real-valued function of a single real variable, and let
$$ u : mathbb{R}^2 to mathbb{R} $$
be a real-valued function of two real variables defined by the formula
$$ u = u(x,y) = xy.$$
Then the function $g = fcirc u$ is a real-valued function of two real variables. The partial derivatives of $g$ can be found via the chain rule:
$$ g_x
= frac{mathrm{d}(fcirc u)}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}
= frac{mathrm{d}f}{mathrm{d}u} cdot y,$$
where $frac{mathrm{d}f}{mathrm{d}u}$ is the derivative of the single variable function $f$. This derivative is not taken with respect to either $x$ or $y$, but instead with respect to the variable of $f$, which we have labeled $u$. Note that as $f$ is a function of a single variable $u$, it makes sense to write $f'(u)$ instead of $frac{mathrm{d}f}{mathrm{d}u}(u)$.

Perhaps an example would help: suppose that $f(u) = cos(u)$, and take $u = xy$. Then
$$ frac{mathrm{d}f}{mathrm{d}x}
= frac{mathrm{df}}{mathrm{d}u} cdot frac{mathrm{d}u}{mathrm{d}x}. tag{1}$$
But
$$ frac{mathrm{d}f}{mathrm{d}u} = frac{mathrm{d}}{mathrm{d}u} cos(u)
= -sin(u). $$
Then, as $u = xy$, it follows that $frac{mathrm{d}f}{mathrm{d}u} = -sin(xy)$. Also note that $frac{mathrm{d}u}{mathrm{d}x} = y$. Substituting these into (1), we conclude that
$$frac{mathrm{d}f}{mathrm{d}x} = -sin(xy) cdot y. $$