Show that $V\otimes V\simeq L(V^*,V^*,\mathbb{R})$

You just need to show that, $L(V^*,V^*,\mathbb{R})$ satisfies the universal property of tensor product.

Consider the map $\Phi :V\times V\to L(V^*,V^*,\mathbb{R}).$

Let, $W$ be a vector space with a bilinear map $B$ from $V\times V$ to $W.$

Suppose, $\{v_i^*\}$ is a basis of $V^*$ and if $v^*=\sum_{i} a^i v_i^{*},w^*=\sum_{j} b^j v_j^*,$ we have, $B'(v^*,w^*)=\sum_{i,j} a^ib^jB'(v_i^*,v_j^*)$

Then, there is a unique linear map $T :L(V^*,V^*,\mathbb{R})\to W$ defined by, $T(B')=\sum_{i,j}a^ib^jB(v_i,v_j)$ such that the diagram commutes.

Thus, from the universal property of tensor product, $V\otimes V\simeq L(V^*,V^*,\mathbb{R})$