I understand why the first term in the solution is ydx. The partial derivative with respect to x is taken of the first term, so x is differentiated to become 1, while the y is held constant, resulting in ((1)y), all multiplied by dx.
However, with regards to the second term in the solution, I do not understand where the x has come from.
If the partial derivative of sin(y) is taken with respect to y, surely the result is (1)(cos(y)), all multiplied by dy. My thinking is that the y term in sin(y) is differentiated to become 1, and this is multiplied by the derivative of the sin term, which is cos(y). This term is then multiplied by dy.
So where has the x term come from?
To get the total differential of $z = f(x,y)$, you need to compute $$ dz = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy. $$ You have in your example $f(x,y) = xy + \sin y$, so you get: \begin{align*} \frac{\partial f}{\partial x} &= \frac{\partial}{\partial x}\left[xy\right] + \frac{\partial}{\partial x}\left[\sin y\right]\\ &= y + 0\\ &= y\\ \frac{\partial f}{\partial y} &= \color{red}{\frac{\partial}{\partial y}\left[xy\right]} + \frac{\partial}{\partial y}\left[\sin y\right]\\ &= \color{red}{x} + \cos y. \end{align*} Thus, the total derivative is $$ dz = ydx + (x + \cos y) dy. $$
Your misunderstanding appears to be in thinking that you only take the partial derivative of one term in the sum with respect to $x$ or $y$, whereas you need to take the partial derivative of the entire sum.