This is Definition 1.4.1 and Lemma 1.4.2 from the book W._Scharlau: Quadratic and Hermitian forms.
4.1. Definition. A vector $x\ne 0$ in a bilinear space $(V,b)$ is called isotropic if $b(x,x) = 0$. Otherwise $x$ is called anisotropic. If $(V,b)$ contains an isotropic vector, then $(V,b)$ is called isotropic as well. We will also say that $(V,b)$ represents zero. Otherwise $(V,b)$ is called anisotropic. A subspace $W$ of $V$ is called totally isotropic if $b(W, W)=0$, that is $b(x,y)=0$ for all $x,y\in W$.
Let $V$ be a vector space and $V^*$ its dual space. On the vector space $V\oplus V^*$ we consider the following symmetric bilinear form \begin{gather*} h=h_V\colon (V\oplus V^*)\times (V\oplus V^*) \to K\\ h((x,f),(y,g))=fy+gx \end{gather*} (One can easily check that $h$ is a symmetric bilinear form.)
4.2. Lemma. (i) $(V\oplus V^*,h_V)$ is regular.
(ii) $V$ and $V^*$ are totally isotropic subspaces.
I found bit confusing how to show $V$ and it's dual is totally isotropic for that I need to show $h(x,y)=0$ for all $x$, $y$ in $V$ and it's dual. Where $h$ is given bilinear symmetric form
You just need to check that $$h((x,0),(y,0))=0 \qquad\text{and}\qquad h((0,f),(0,g))=0$$ for any $x,y\in V$ and $f,g\in V^*$.
The first equation shows that $V$ is a totally isotopic subspace of $V\oplus V^*$. The second equation shows that this is true for $V^*$. (Notice that we identify $V$ with $V\times\{0\}$ and $V^*$ with $\{0\}\times V^*$.)
I will remind you (look at the Definition 4.1) that to show that something is a totally isotropic subspace you only need $b(x,y)$ for $x$ and $y$ from this subspace. In this case you are applying this to the bilinear for $h$ and to the subspaces $V$ and $V^*$.