Determine the partial derivatives with respect to $x$ and $y$ of the function
$$z(x,y) = \dfrac{3x - y}{2y+3}.$$
When determining the partial derivative with respect to $x$, it is useful to write $z(x,y)$ as
$$z(x,y) = \dfrac{3x}{2y+3} - \dfrac{y}{2y+3} = \dfrac{1}{2y+3} \cdot 3x - \dfrac{y}{2y+3},$$
such that $z(x,y)$ can be written as
$$z(x,y) = c \cdot u(x,y) + d \qquad \text{with} \qquad c = \dfrac{1}{2y+3}, \qquad d=- \dfrac{y}{2y+3} \qquad \text{and} \qquad u(x,y) = 3x.$$
Since we determine the partial derivative with respect to $x$, we can consider $y$ as a constant; this means that for the determination of the derivative with respect to $x$ both $c$ and $d$ can be seen as constants. By the use of the scalar product rule and the sum rule we now find the partial derivative of $z(x,y)$:
$$z'_x(x,y) = c \cdot u'_x(x,y) + 0 = \dfrac{1}{2y+3} \cdot 3 + 0 = \dfrac{3}{2y+3}.$$
When determining the partial derivative with respect to $y$, then we can write $z(x,y)$ as the quotient of two functions:
$$z(x,y) = \dfrac{w(x,y)}{v(x,y) }\qquad \text{with} \qquad w(x,y) = 3x-y \qquad \text{and} \qquad v(x,y) = 2y+3.$$
In order to apply the quotient rule we need the partial derivatives of $w(x,y)$ and $v(x,y)$ with respect to $y$. Note that $x$ is a constant; it holds that
$$w'_y(x,y) = 0 - 1 = -1 \qquad \text{and} \qquad v'_y(x,y) = 2 + 0 = 2.$$
According to the quotient rule we get:
$$z'_y(x,y) = \dfrac{w'_x(x,y)v(x,y) - w(x,y)v'_x(x,y)}{\big(v(x,y)\big)^2} = \dfrac{-1\cdot(2y+3) - (3x-y)\cdot2}{(2y+3)^2}= \dfrac{-2y-3 - 6x+2y}{(2y+3)^2} }= \dfrac{-6x-3}{(2y+3)^2}.$$
$$z(x,y) = \dfrac{3x - y}{2y+3}.$$
When determining the partial derivative with respect to $x$, it is useful to write $z(x,y)$ as
$$z(x,y) = \dfrac{3x}{2y+3} - \dfrac{y}{2y+3} = \dfrac{1}{2y+3} \cdot 3x - \dfrac{y}{2y+3},$$
such that $z(x,y)$ can be written as
$$z(x,y) = c \cdot u(x,y) + d \qquad \text{with} \qquad c = \dfrac{1}{2y+3}, \qquad d=- \dfrac{y}{2y+3} \qquad \text{and} \qquad u(x,y) = 3x.$$
Since we determine the partial derivative with respect to $x$, we can consider $y$ as a constant; this means that for the determination of the derivative with respect to $x$ both $c$ and $d$ can be seen as constants. By the use of the scalar product rule and the sum rule we now find the partial derivative of $z(x,y)$:
$$z'_x(x,y) = c \cdot u'_x(x,y) + 0 = \dfrac{1}{2y+3} \cdot 3 + 0 = \dfrac{3}{2y+3}.$$
When determining the partial derivative with respect to $y$, then we can write $z(x,y)$ as the quotient of two functions:
$$z(x,y) = \dfrac{w(x,y)}{v(x,y) }\qquad \text{with} \qquad w(x,y) = 3x-y \qquad \text{and} \qquad v(x,y) = 2y+3.$$
In order to apply the quotient rule we need the partial derivatives of $w(x,y)$ and $v(x,y)$ with respect to $y$. Note that $x$ is a constant; it holds that
$$w'_y(x,y) = 0 - 1 = -1 \qquad \text{and} \qquad v'_y(x,y) = 2 + 0 = 2.$$
According to the quotient rule we get:
$$z'_y(x,y) = \dfrac{w'_x(x,y)v(x,y) - w(x,y)v'_x(x,y)}{\big(v(x,y)\big)^2} = \dfrac{-1\cdot(2y+3) - (3x-y)\cdot2}{(2y+3)^2}= \dfrac{-2y-3 - 6x+2y}{(2y+3)^2} }= \dfrac{-6x-3}{(2y+3)^2}.$$