Let $v \equiv F(y, z)$. The partial derivatives of $F$ are $$F\_1 \equiv \frac{\partial F(y,z)}{\partial y} = \frac{H(v)}{H(y)},$$ $$F\_2 \equiv \frac{\partial F(y,z)}{\partial z} = r\frac{H(v)}{H(z)},$$ where $H$ is an arbitrary function. This is taken from the book [Probability Theory](http://rads.stackoverflow.com/amzn/click/0521592712) by E.T. Jaynes. 

The author goes on saying that the relation $ \mathrm{d}v = \mathrm{d}F(y, z) = F\_1\mathrm{d}y + F\_2\mathrm{d}z$ takes the form of 
$$\frac{\mathrm{d}v}{H(v)} = \frac{\mathrm{d}y}{H(y)} + r\frac{\mathrm{d}z}{H(z)}$$

or, on integration
$$w[F(y, z)] = w(v) = w(y)w^r(z),$$

where
$$w(x) \equiv exp\left\{\int^x \frac{\mathrm{d}x}{H(x)}\right\}.$$

I did undergraduate calculus; however, we didn't do this kind of differential manipulation (expansion of $\mathrm{d}v$) and I find it quite confusing. Is there a way to arrive at the same result using "standard" steps like integration by substitution etc.?