Let $f:\mathbb R^3\to \mathbb R$ be a function such that $$f(\vec p)=h(g_1(\vec p),g_2(\vec p),g_3(\vec p))$$
Where $h=x^2-yz$ and $g_1=x+y,g_2=y^2,g_3=x+z$
$1.$way :Since $f(x,y,z)=h(x+y,y^2,x+z)=(x+y)^2-y^2(x+z)$ it is easy to find $\dfrac{\partial}{\partial x}f=2(x+y)-y^2$
But I want to use the formal definition without putting and taking partial wrt $x$
$2.$ way:(I couldnot understand, it is where I am having confusion) $\dfrac\partial{\partial x} f\\=\dfrac{\partial f}{\partial g_1}\dfrac{\partial g_1}{\partial x}+\dfrac{\partial f}{\partial g_2}\dfrac{\partial g_2}{\partial x}+\dfrac{\partial f}{\partial g_3}\dfrac{\partial g_3}{\partial x}\\=\dfrac{\partial f}{\partial (x+y)}\underbrace{\dfrac{\partial (x+y)}{\partial x}}_1+\dfrac{\partial f}{\partial (y^2)}\underbrace{\dfrac{\partial y^2}{\partial x}}_0+\dfrac{\partial f}{\partial (x+z)}\underbrace{\dfrac{\partial (x+z)}{\partial x}}_1$
Now I am stuck
You've done most of the work.Since $\frac{\partial f}{\partial g_1}=2g_1,\,\frac{\partial f}{\partial g_2}=-g_3,\,\frac{\partial f}{\partial g_3}=-g_2$,$$\partial_xf=2(x+y)1-(x+z)0-y^21=2(x+y)-y^2.$$