#### Answer

False

#### Work Step by Step

Consider $x=f(t)=(t-1)^{3}$ and $y=g(t)=(t-1)^{2}$
Thus
$g'(t)=2(t-1)$
Now put $t=1$
$g'(1)=2(1-1)=0$ which is not equal to $1$ as per the given statement , that is, $g'(1)\ne 1$ but it has vertical tangent at $t=1$
Hence , the statement is false.