We put ourselves in a suitable common domain, i.e. the finite particle vectors. Then $$\bigl(a^*(f)\Psi\bigr)_n(X_n)=\frac{1}{\sqrt{n}}\sum_{j=1}^n f(x_j)\Psi_{n-1}(X_n\setminus{x_j})\\ \bigl(a(f)\Psi\bigr)_n(X_n)=\sqrt{n+1}\int \bar{f}(x)\Psi_{n+1}(x,X_n)dx$$ Hence by definition $$\bigl(a(g)a^*(f)\Psi\bigr)_n(X_n)=\sum_{j=1}^{n+1}\int \bar{g}(x_1)f(x_j)\Psi_n(X_{n+1}\setminus x_j)\\\bigl(a^*(f)a(g)\Psi\bigr)_n(X_n)=\sum_{j=1}^{n}\int \bar{g}(x)f(x_j)\Psi_n(x,X_{n}\setminus x_j)\; .$$ The result immediately follows subtracting.

For bosons: We put ourselves in a suitable common domain, i.e. the finite particle vectors. Then $$\bigl(a^*(f)\Psi\bigr)_n(X_n)=\frac{1}{\sqrt{n}}\sum_{j=1}^n f(x_j)\Psi_{n-1}(X_n\setminus{x_j})\\ \bigl(a(f)\Psi\bigr)_n(X_n)=\sqrt{n+1}\int \bar{f}(x)\Psi_{n+1}(x,X_n)dx$$ Hence by definition $$\bigl(a(g)a^*(f)\Psi\bigr)_n(X_n)=\sum_{j=1}^{n+1}\int \bar{g}(x_1)f(x_j)\Psi_n(X_{n+1}\setminus x_j)\\\bigl(a^*(f)a(g)\Psi\bigr)_n(X_n)=\sum_{j=1}^{n}\int \bar{g}(x)f(x_j)\Psi_n(x,X_{n}\setminus x_j)\; .$$ The result immediately follows subtracting. For fermions is similar.