Differentiation chain rule If $f(tx,ty,tz) = t^nf(x,y,z)$ then in my lecture notes it says differentiating this equation with respect to $t$ gives:
$$x\frac{df}{dx} + y\frac{df}{dy} + z\frac{df}{dz} = nt^{n-1}f(x,y,z)$$ 
But why isn't it:
$$x\frac{df}{d(tx)} + y\frac{df}{d(ty)} + z\frac{df}{d(tz)} = nt^{n-1}f(x,y,z)\ ?$$
 A: Suppose $f(tx,ty,tz)=t^nf(x,y,z)$. Consider the chain rule:
$$ \frac{d}{dt} f(x_1(t),x_2(t),x_3(t)) = \frac{\partial f}{\partial x_1}(x_1(t),x_2(t),x_3(t))\frac{dx_1}{dt}+\frac{\partial f}{\partial x_2}(x_1(t),x_2(t),x_3(t))\frac{dx_2}{dt}+\frac{\partial f}{\partial x_3}(x_1(t),x_2(t),x_3(t))\frac{dx_3}{dt} $$
I am being really pedantic here, but the question raised involves this issue. The partial derivatives are properly understood to be evaluated at the inside function which happens to be $(x_1(t),x_2(t),x_3(t))$ in this context. Returning to the given identity, identify $x_1=tx$ and $x_2=ty$ and $x_3=tz$ where presumably $x,y,z$ are not functions of time (if they were we'd obtain a different identity). Note:
$$ \frac{dx_1}{dt} = x, \ \ \frac{dx_2}{dt} = y, \ \ , \frac{dx_3}{dt} = z $$
Consequently,
$$ \frac{d}{dt} f(x_1(t),x_2(t),x_3(t)) = x\frac{\partial f}{\partial x_1}(x_1(t),x_2(t),x_3(t))+y\frac{\partial f}{\partial x_2}(x_1(t),x_2(t),x_3(t))+z\frac{\partial f}{\partial x_3}(x_1(t),x_2(t),x_3(t)) $$
which means,
$$ \frac{d}{dt} f(tx,ty,tz) = x\frac{\partial f}{\partial x_1}(tx,ty,tz)+y\frac{\partial f}{\partial x_2}(tx,ty,tz)+z\frac{\partial f}{\partial x_3}(tx,ty,tz) $$
So, I tend to think your book takes some liberty with notation.
A: Your thoughts are mostly correct, but just now written clearly.
Normally, a function in three variables is denoted $f(x, y, z)$, and the partial derivatives are $\frac{\partial f}{\partial x}$, $\frac{\partial f}{\partial y}$, $\frac{\partial f}{\partial z}$, what they mean is the partial derivative with respect to each variable.
To interpret $f(tx, ty, tz)$ nicely, we shall write $f$ as a function $f(x_1, x_2, x_3)$, and $x_1(t) = tx, x_2(t) = ty, x_3(t) = tz$, so chain rule gives
$$
\frac{d}{dt}f(x_1(t), x_2(t), x_3(t)) = \frac{dx_1}{dt}\frac{\partial f}{\partial x_1} + \frac{dx_2}{dt}\frac{\partial f}{\partial x_2} + \frac{dx_3}{dt}\frac{\partial f}{\partial x_3}.
$$
So you are confused with $x, y, z$ denoting the first, second, third terms of $f$, and the fact that are used as other variables again.
A: The notes is alright but not clear enough, you are in the right direction as well, but your notation is not the best way to illustrate it. 
The $\partial f/\partial x$ in your notes is rather taking derivative with respect to the first variable than taking derivative with respect to $x$. 
Try to think in this way. Say you have a function with three slots to fill (notice this function satisfy the relation you gave, so it may serve as an example):
$$
f(\square, \triangle,\bigcirc) = \square^n + \triangle^n + \bigcirc^n.\tag{1}
$$
Now we fill the first variable by $x(t)$, the other two untouched then:
$$
\frac{df}{dt} = \frac{\partial }{\partial \square}f(\square, \triangle,\bigcirc) \Bigg\vert_{\square = x(t)} \cdot \frac{d x(t)}{dt},
$$
the first term is taking derivative with respect to the first variable evaluated at $x(t)$. Taking (1) as an example: let $x(t) = tx$
$$
\frac{df}{dt} = \frac{\partial }{\partial \square}f(\square, \triangle,\bigcirc) \Bigg\vert_{\square = tx} \cdot x = n \square^{n-1}\big|_{\square = tx} \cdot x = nt^{n-1}x^n,
$$
If we evaluate directly 
$$
f(tx, \triangle,\bigcirc) = t^n x^n + \triangle^n + \bigcirc^n,
$$
you will get the same result as chain rule.
Keeping this in mind your formula can be written in a least ambiguous way: for $f(u,v,w)$
$$x\frac{\partial f}{\partial u}\Bigg\vert_{u = tx} + y\frac{\partial f}{\partial v} \Bigg\vert_{v = ty} + z\frac{\partial f}{\partial w}\Bigg\vert_{w = tz}  = nt^{n-1}f(x,y,z).$$ 
