Show that $\langle Tx,Ty\rangle_1=\langle y, x\rangle$ defines an inner product on $H\mbox{*}$ that induces a norm on $H\mbox{*}$ The correspondence $T:H\rightarrow H\mbox{*}$ by $y\mapsto f_y$, and $y\in H$ and $T$ is a conjugate-linear isometry. We call $H\mbox{*}$ the dual space. We know that something is conjugate-linear if $$T(\lambda u+\mu v)=\bar{\lambda} Tu+\bar{\mu} Tv$$ for all $u\in X,v\in Y$ and $\lambda,\mu\in\mathbb{F}$. And isometry being $||Tx||=||x||$
We want to show that $$\langle Tx,Ty\rangle_1=\langle y,x\rangle$$ defines an inner product on $H\mbox{*}$ that induces a norm on $H\mbox{*}$.
I know that I am supposed to show that $$||x||=0\iff x=0$$
$$\text{Triangle Inequality}$$
$$||\alpha x||=|\alpha|||x||$$
But I am struggling with the exact steps to do this.
I will post what I have in a little if you would be more open to checking what I have.
This is what I was thinking but for some reason I just felt like it wasn't quite right.
[(a)][$||x||=0 \iff x=0$]:
\begin{align*}
    ||Tx||&=0\\
    ||Tx||^2&=\\
    \langle Tx,Tx\rangle_1&=\\
    \langle x,x\rangle&=0\iff x=0\\
\end{align*}
[(b)][$||\alpha x||=|\alpha|||x||$]:
\begin{align*}
    ||T\alpha x||&=||\bar{\alpha}Tx||\\
    &=||\bar{\alpha} x||\\
    &=|\alpha|||x||\\
    &=|\alpha|||Tx||
\end{align*}
[(c)][Triangle inequality]:
\begin{align*}
    ||Tx+Ty||&=||x+y||\\
    &\leq||x||+||y||\\
    &=||Tx||+||Ty||
\end{align*}

Your attempt does not show what it should.
You are supposed to show that $\langle Tx,T_y\rangle_1:= \langle y,x\rangle$ defines an inner product. Nowhere in your attempt you even speak about an inner product.
