optim function from R in C++ with the Rcpp package

Question

I have a data that look like this (few k of observations)

df <- data.frame(age1 = c(10, 20, 30), age2 = c(20, 30, 40), age3 = c(30, 40, 50), 
                 dia1 = c(15, 20, 25), dia2 = c(20, 25, 30), dia3 = c(25, 30, 35))

I want to calculate coefficients of the growth functions for each row in the data frame and I use optim function from R. It takes some time to run it thus I have tried to develop Rcpp function that do the same thing

library(Rcpp)
cppFunction('
            List
            computeIndex(
            const NumericVector dia1,
            const NumericVector dia2,
            const NumericVector dia3,
            const NumericVector age1,
            const NumericVector age2,
            const NumericVector age3)
            {

            int n = dia1.size();
            NumericVector  coef1(n), coef2(n), coef3(n), age(n), d(n), f(n);

            for(int i = 0; i < n; ++i)
            {
             age[i] = (age1[i], age2[i], age3[i]);
             d[i] = (dia1[i], dia2[i], dia3[i]);


              f[i] = function(param) sum((d[i] - (param[1] *  exp(-param[2] * exp(-param[3] * age[i]))))^2);

              coef1[i] = optim(c(0, 0, 0), f, method = "BFGS")$par[1];
              coef2[i] = optim(c(0, 0, 0), f, method = "BFGS")$par[2];
              coef3[i] = optim(c(0, 0, 0), f, method = "BFGS")$par[3];
            }

            return List::create(Named("coef1") = coef1,
            Named("coef2") = coef2, Named("coef3") = coef3);
            }
            ')

Unfortunately have no idea how to use the optim from R in the Rcpp function. I read the post concerning issue but don't understand the solution have been proposed there,

The R code that work is

library(dplyr)
df1 <- df  %>% rowwise() %>%
  do({
    age <- c(.$age1, .$age2, .$age3)
    d <- c(.$dia1, .$dia2, .$dia3)
    f <- function(param) sum((d - (param[1] *  exp(-param[2] * exp(-param[3] * age))))^2)
    data.frame(., coef1 = optim(c(0, 0, 0), f, method = "BFGS")$par[1], coef2 = optim(c(0, 0, 0), f, method = "BFGS")$par[2], 
               coef3 = optim(c(0, 0, 0), f, method = "BFGS")$par[3])
  } )

Answer is

   age1  age2  age3  dia1  dia2  dia3    coef1    coef2      coef3
* <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>    <dbl>    <dbl>      <dbl>
1    10    20    30    15    20    25 45.46349 1.514253 0.03086497
2    20    30    40    20    25    30 41.02628 1.647884 0.04097674
3    30    40    50    25    30    35 45.13945 2.078568 0.04149476