Show that $\mathrm{d}S=\frac{1}{T}\,\mathrm{d}U+\frac{1}{T}\,P\,\mathrm{d}V-\frac{1}{T}\,\mu\,\mathrm{d}N$ I need help to show that \begin{align*}\mathrm{d}S &=\left(\frac{\partial S}{\partial U}\right)_{V,N}\mathrm{d}U+\left(\frac{\partial S}{\partial V}\right)_{U,N}\mathrm{d}V+\left(\frac{\partial S}{\partial N}\right)_{V,U}dN\tag{1}\\&=\frac{1}{T}\mathrm{d}U+\frac{1}{T}P\mathrm{d}V-\frac{1}{T}\mu\mathrm{d}N\tag{2}\end{align*}
where $U$ is the Internal Energy of the system; $S$ is the Entropy of the system; $N$ is the Number of Particles in the system; $V$ is the Volume of the system; $P$ is the systems' Pressure; $T$ is the absolute (thermodynamic) temperature of the system and $\mu$ is the Chemical Potential of the system.
I know that the coefficients of $\mathrm{d}U$,$\,\mathrm{d}V$ and $\mathrm{d}N$ must match for equations $(1)$ and $(2)$ ie.
$$\left(\frac{\partial S}{\partial U}\right)_{V,N}=\frac{1}{T}\tag{A}$$
$$\left(\frac{\partial S}{\partial V}\right)_{U,N}=\frac{1}{T}P\tag{B}$$
$$\left(\frac{\partial S}{\partial N}\right)_{V,U}=-\frac{1}{T}\mu\tag{C}$$
But I simply have no idea how to show $(\mathrm{A})$, $(\mathrm{B})$ and $(\mathrm{C})$. So this means I am stuck at the very beginning and hence cannot show my attempt at providing a solution (reason for question closure).

For context I have added the pages of my text that shows the equivalence of equations $(1)$ and $(2)$:





Could someone please help me show that 
\begin{align*}&\left(\frac{\partial S}{\partial U}\right)_{V,N}\mathrm{d}U+\left(\frac{\partial S}{\partial V}\right)_{U,N}\mathrm{d}V+\left(\frac{\partial S}{\partial N}\right)_{V,U}dN\\&=\frac{1}{T}\mathrm{d}U+\frac{1}{T}P\mathrm{d}V-\frac{1}{T}\mu\mathrm{d}N\end{align*}
Any hints or tips are greatly appreciated. Thanks.
 A: $$\left(\frac{\partial S}{\partial U}\right)_{V,N}=\frac{1}{T}\tag{A}$$
Is defined as an expression for temperature and is not derived.
Once they teach you entropy, they use it to tighten the definition of temperature. 
$$\left(\frac{\partial S}{\partial V}\right)_{U,N}=\frac{1}{T}P\tag{B}$$
is also introduced by reasoned argument, as the imposed  definition of pressure, (when you get the $P$  on its own),  rather than any derivation and I am inclined to believe this is because it might involve a Legrande transformation, which would take to long to explain,  as well being slightly off-topic.
$$\left(\frac{\partial S}{\partial N}\right)_{V,U}=\frac{1}{T}\mu\tag{C}$$
The final expression involves extending the thermodynamic equation to include "chemical work" and you get it by 
$$\mathrm{d}U= T\mathrm{d}S -P\mathrm{d}V + \mu \mathrm{d}N$$
Now $U, S, V, N $  are all capable of change in the above  equation.
So imagine we hold the variables  $U,S $ fixed
Such that $$0 = T\mathrm{d}S + \mu \mathrm{d}N$$   
leads to $$ \mu = -T \left(\frac{\partial S}{\partial N}\right)_{V,U}$$
which leads to $$\left(\frac{\partial S}{\partial N}\right)_{V,U}=-\frac{1}{T}\mu\tag{C}$$
