Prob. 7, Sec. 3.8 in Erwine Kreyszig's INTRODUCTORY FUNCTIONAL ANALYSIS WITH APPLICATIONS: The dual space of a Hilbert space is a Hilbert space. Here's Prob. 7, Sec. 3.8 in Introductory Functional Analysis With Applications by Erwine Kreyszig: 

Show that the dual space $H^\prime$ of a Hilbert space $H$ is a Hilbert space with inner product $\langle \ \cdot \ , \ \cdot \ \rangle_1$ defined by 
  $$\langle f_z, f_v \rangle_1 \ = \ \overline{\langle z, v \rangle} \ = \ \langle v, z \rangle,$$
  where $f_z(x) = \langle x, z \rangle$ for all $x \in X$. 

Now I know that each bounded linear functional $f \in H^\prime$ can be written as $f = f_z$, for a unique $z \in H$ with $\Vert z \Vert = \Vert f \Vert$. 
The sum of two bounded linear functionals on any normed space is again a bounded linear functional, and so is the scalar multiple of any bounded linear functional.  
Moreover, for each $z \in H$ and for each $w \in H$, we have 
$$\left( f_z + f_w \right) (x) = f_z(x) + f_w(x) = \langle x, z \rangle + \langle x, w \rangle = \langle x, z+w \rangle = f_{z+w}(x)$$ 
for all $x \in H$. So $f_{z+w} = f_z + f_w$. 
For $z \in H$ and for any scalar $\alpha$, we have 
$$f_{\alpha z} (x) = \langle x, \alpha z \rangle = \overline{\alpha} \langle x, z \rangle = \overline{\alpha} f_z (x) \ \mbox{ for all } \ x \in H.$$
So, $f_{\alpha z} = \overline{\alpha } f_z$. 
Thus the mapping $z \mapsto f_z$ of $H$ into $H^\prime$ is surjective, isometric, and conjugate linear. 
Moreover, the set of all bounded linear functionals on any normed space is itself a normed space, rather a Banach space. 
But how to show that the inner product given by Kreyszig is the (only) natural one (i.e. that which fits into what we already know from the normed space theory)? 
 A: Let me clarify what the question is asking, since based on your comments you actually already know how to solve it.
As you know, for any complex normed space $(X, \|\cdot\|_X)$, the dual space $X'$ is the complex vector space of all continuous linear functionals $f : X \to \mathbb{C}$, with pointwise addition and scalar multiplication.  We usually equip $X'$ with the "operator norm" $\|\cdot\|_{X'}$ defined by
$$\|f\|_{X'} = \sup_{x \in X, \|x\|_X = 1} |f(x)|.$$
Now in case $X$ is a Hilbert space $(H, \langle \cdot, \cdot \rangle_H)$, we know from the Riesz representation theorem that every $f \in H'$ is of the form $f_z$ for some $z \in H$, where $f_z$ is defined by $f_z(v) = \langle v, z \rangle$.  This suggests constructing an inner product on the complex vector space $H'$: Kreyszig calls this new inner product $\langle \cdot, \cdot \rangle_1$, and it is defined by $\langle f_z, f_v \rangle_1 = \langle v, z \rangle_H$.  You should verify that $\langle \cdot, \cdot \rangle_1$ is an inner product (sesquilinear, Hermitian, positive definite) but this is very easy.
Now, on the complex vector space $H'$ we have a norm $\|\cdot\|_{H'}$ and an inner product $\langle \cdot, \cdot \rangle_1$.  The question is asking you to show that the inner product $\langle \cdot, \cdot \rangle_1$ induces the norm $\|\cdot\|_{H'}$.  That is, you must show that $\|f\|_{H'} = \sqrt{\langle f, f \rangle_1}$ for every $f \in H'$.
If you unwind this, it comes down to showing something like the following: for every $z \in H$, $$\|z\|_H = \sup_{v \in H, \|v\|=1} |\langle v,z \rangle|.$$
Prove this by proving two inequalities.  For one of them, use Cauchy-Schwarz.  For the other, consider an appropriate choice for $v$.
You only have to show this for the specific inner product $\langle \cdot, \cdot \rangle_1$ given in the question, not for an arbitrary inner product on $H'$.  The question is not asking you to show that every inner product on $H'$ induces the norm $\|\cdot\|_{H'}$.  This is trivially false; as a silly example, consider the inner product $\langle \cdot, \cdot\rangle_2$ defined by $\langle f_z, f_v \rangle_2 = 2 \langle v, z \rangle_H$.  If you use the axiom of choice, you can produce other inner products on $H'$ that induce completely different (and uninteresting) topologies.
