Let $E$ be a elliptic curve that is , genus $1$ smooth curve with base point. And suppose that the map $φ$, which sends $E$ to $E'$, is birational. Birational means there are two rational maps on both direction which composites identity.
Then, can we say that $E'$ is smooth?
No, this is distinctly untrue. Indeed, finding a birational map to a smooth curve is often how we deal with singular curves. For a specific counterexample consider $$C : Y^2Z^2 = X^4 + Z^4$$ with the base point $O = [0:1:1]$. Note that $C$ is a singular curve with singular point $[0:1:0]$, the point at infinity.
Then $C$ is birational to the (smooth) elliptic curve $$E : Y^2Z = X^3 - 4XZ^2$$ with the map given by $$\phi : [X,Y,Z] \mapsto [2XYZ + 2XZ^2, 4YZ^2 + 4Z^3,X^3]$$
The inverse of $\phi$ is given by $$ [X,Y,Z] \mapsto [2YZ, X^2 + 4Z^2,X^2 - 4Z^2]$$